Assumptions
Our Bitcoin pricing model is derived from three major assumptions.
Assumption 1: Value (purchasing power) cannot be created or destroyed and is finite in
substance. Hence the entire financial eco-system is predicated on ZERO-sum game dynamics.
For example, for one asset to increase in market capitalisation there needs to be another asset
(or multiple assets) that equivalently decrease in market capitalisation.
Assumption 2: In short time horizons, the aggregate change of value for all assets within
the financial ecosystem must equate to ZERO in order to maintain finite substance condition
in assumption 1. Changes in the value of Bitcoin, can only be occurring via some form of
substitution phenomenon. Substitution itself must be bi-directional representing the rate of
value absorption/dispersion from alternative asset within its own class as well as assets outside
of its class.
Assumption 3 : Substitution is driven by Markowitz portfolio optimisations from which
multiple rational economic agents are competing to maximise individual investor utility whilst
minimising their individual portfolio risk as close to ZERO as physically possible. Assumption
3 suggests economic agents tend to embrace Bitcoin as a superior store of value (especially
considering longer time horizons) which can combat inflation and provide a unique hedge against
systemic risk in the traditional financial system and/or traditional currencies
Representation of a New Asset Class via Substitution
First, consider the case of 2 assets. Then, the Wronskian of the functions M
1
and M
2
will be
W (M
1
, M
2
) =
M
1
M
2
M
′
1
M
′
2
= M
1
· M
′
2
− M
′
1
· M
2
.
On the other hand, we know that
∂ (M
1
+ M
2
)
∂t
= M
′
1
+ M
′
2
= 0, (1)
leading to
M
′
1
= −M
′
2
.
Substituting this into the expression for W (M
1
, M
2
), we obtain that
W (M
1
, M
2
) = M
1
· M
′
2
− (−M
′
2
) · M
2
= M
1
· M
′
2
+ M
′
2
· M
2
= M
′
2
· (M
1
+ M
2
).
Apparently, W (M
1
, M
2
) = 0 at least for some t, providing that M
1
and M
2
are linearly inde-
pendent functions of time. However, it is important to note that in view of the equality (1),
M
′
1
and M
′
2
are not linearly independent.
The linear independence of M
1
and M
2
means that any new asset M
3
in the vector space
with basis functions M
1
and M
2
can be represented as
M
3
= ω
1
· M
1
+ ω
2
· M
2
,
where ω
1
and ω
2
are not simultaneously 0.
Evidently, for any ω
3
= −1, we can write
ω
1
· M
1
+ ω
2
· M
2
+ ω
3
M
3
= 0
for the above ω
1
and ω
2
. Therefore, M
1
, M
2
and M
3
are linearly independent functions forming a
basis for a new, higher-dimensional vector space, in which, any new asset M
4
can be represented
as
M
4
= ω
1
· M
1
+ ω
2
· M
2
+ ω
3
M
3
.
In a similar fashion, one can show that in the n-dimensional space with basis functions M
1
,
M
2
, . . ., M
n
, the new asset M
BTC
can be represented as
M
BTC
=
n
X
k=1
ω
k
· M
k
(2)
with ω
k
= −1 which is satisfied in our case since all ω
k
> 0.
Hereinafter, M
k
is regarded as the total capitalization of Market k, and ω
k
– as the substi-
tution of Market k.
The Governing Equation
We start with the equation of exchange below:
P
BTC
· Q
BTC
= M
BTC
· V
BTC
,
where P
BTC
is the price of asset, Q
BTC
is the output power, M
BTC
is the market capitalization,
and V
BTC
is the velocity of the asset class.
We consider the general case when Q
BTC
, M
BTC
and V
BTC
are functions of time variable t,
i.e., in what follows, Q
BTC
= Q
BTC
(t), M
BTC
= M
BTC
(t) and V
BTC
= V
BTC
(t).
Evidently, taking natural logarithm in both sides of the previous equality, we derive
ln P
BTC
+ ln Q
BTC
= ln M
BTC
+ ln V
BTC
.
Differentiating both sides of this expression with respect to t, we obtain
∂
∂t
[ln P
BTC
+ ln Q
BTC
− (ln M
BTC
+ ln V
BTC
)] = 0.
or
∂ ln P
BTC
∂t
=
∂
∂t
[ln M
BTC
+ ln V
BTC
− ln Q
BTC
] .
(3)
For the sake of simplicity, we denote
∂ ln P
BTC
∂t
= Π
BTC
.
Therefore:
Π
BTC
=
∂
∂t
[ln M
BTC
+ ln V
BTC
− ln Q
BTC
] .
A Particular Model for Capitalization
We now assume that
M
k
= P
k
· U
k
, (4)
where P
k
is the price and U
k
is the unit of the kth asset.
Then,
Π
BTC
=
∂
∂t
"
ln
n
X
k=1
ω
k
· P
k
· U
k
!
+ ln V
BTC
− ln Q
BTC
#
.
(5)
A Particular Model for Substitution
In this section, we consider a specific model for the absorption rate ω
k
as follows
ω
k
=
U
s
k
U
k
(6)
for k = 1, 2, . . . , n and where U
sk
is the substitution unit of the kth asset. Substituting it into
(5), we derive
Π
BTC
=
∂
∂t
"
ln
n
X
k=1
P
k
· U
k
·
U
s
k
U
k
!
+ ln V
BTC
− ln Q
BTC
#
=
=
∂
∂t
"
ln
n
X
k=1
P
k
· U
s
k
!
+ ln V
BTC
− ln Q
BTC
#
.
A Particular Model for Velocity
Consider the following model for V
BTC
:
V
BTC
=
1
m
m
X
j=1
T
′
j
, (7)
where T
j
represents the transactions. Substituting it into the final expression of Π
BTC
leads us
to
Π
BTC
=
∂
∂t
"
ln
n
X
k=1
P
k
· U
s
k
!
+ ln
1
m
m
X
j=1
T
′
j
!
− ln Q
BTC
#
.
A Particular Model for Output
Now, we assume that
Q
BTC
=
b · h
d
, (8)
where b, h and d are time-dependent production parameters. Therefore, for Π
BTC
, we will have
Π
BTC
=
∂
∂t
"
ln
n
X
k=1
P
k
· U
s
k
!
+ ln
1
m
m
X
j=1
T
′
j
!
− ln
b · h
d
#
=
=
∂
∂t
"
ln
n
X
k=1
P
k
· U
s
k
!
+ ln
1
m
m
X
j=1
T
′
j
!
− (ln b + ln h − ln d)
#
=
=
∂
∂t
"
ln
n
X
k=1
P
k
· U
s
k
!
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d
#
.
The Absorption Consideration
Assume that
U
s
k
= α
k
· R
k
, (9)
where α
k
is the absorption rate of the market k, R
k
, k = 1, 2, ·n, are time-dependent.
Then, Π
BTC
will obtain the following form:
Π
BTC
=
∂
∂t
"
ln
n
X
k=1
P
k
· α
k
· R
k
!
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d
#
.
Hence the Zero Theorem governing equation is:
Π
BTC
=
∂
∂t
"
ln
n
X
k=1
α
k
· P
k
· R
k
!
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d
#
.
Investigating Alpha
In this section, we are going to investigate the time-dependent behavior of the absorption rate
(α) in two principal cases: when the absorption rate is assumed to be the same for all markets
and when it is unique for each market.
The Case of Market Specific Alpha (α
k
)
In this case, we derive a single equation for all α
k
:
n
X
k=1
P
k
· α
k
· R
k
=
P
BTC
· b · h
d ·
m
X
j=1
T
′
j
. (10)
Note that (10) is a single functional equation with respect to n unknowns α
1
, α
2
, . . ., α
n
.
Apparently, unlike the previous case, it can not be solved exactly for continuous t. Therefore,
we need to consider (10) at discrete values of t where the data measurements are made.
In order to describe the algorithm of determination of α
1
, α
2
, . . ., α
n
, let us denote
F (α
1
, α
2
, . . . , α
n
; t) =
n
X
k=1
P
k
· α
k
· R
k
and
S (t) =
P
BTC
· b · h
d ·
m
X
j=1
T
′
j
.
Then, (10) can be written as
F (α
1
, α
2
, . . . , α
n
; t) = S (t) . (11)
It is important to recognize that in (11), α
1
, α
2
, . . ., α
n
, P
BTC
, b, h, d and T
′
j
are all functions
of time.
Now assume that the raw data measurements have been made at given instances t
i
, i =
1, 2, . . . , N. For example, when data are measured on daily basis, then t
1
, t
2
, . . ., t
N
represent
days. Evaluating both sides of (11) at instances t
i
, i = 1, 2, . . . , N, we will have
F (α
1
(t
1
) , α
2
(t
1
) , . . . , α
n
(t
1
) ; t
1
) = S (t
1
) ,
F (α
1
(t
2
) , α
2
(t
2
) , . . . , α
n
(t
2
) ; t
2
) = S (t
2
) ,
.
.
.
F (α
1
(t
N
) , α
2
(t
N
) , . . . , α
n
(t
N
) ; t
N
) = S (t
N
) .
(12)
Note that (12) is linear with respect to each α
k
(t
i
), k = 1, 2, . . . , n, i = 1, 2, . . . , N. Nonetheless,
(12) contains N equations with respect to n · N unknowns. At this, N ≫ n and the system
is under-determined. Therefore, direct methods can not be applied for solving (12). On the
other hand, in order to determined unknowns α
k
(t
i
), k = 1, 2, . . . , n, i = 1, 2, . . . , N, we can
apply either efficient numerical methods of linear programming, choosing, e.g., P
BTC
as a cost
function, or we can solve the equivalent problem of numerical minimization
||F (α
1
(t
i
) , α
2
(t
i
) , . . . , α
n
(t
i
) ; t
i
) − S (t
i
)|| → min
0≤α
k
(t
i
)≤1
, k = 1, 2, . . . , n i = 1, 2, . . . , N.
Here, ||·|| denotes an appropriate norm. In this case, it can be the l
2
-norm. Then, the problem
formulation will be: determine the solution to the following numerical minimization problem:
|F (α
1
(t
i
) , α
2
(t
i
) , . . . , α
n
(t
i
) ; t
i
) − S (t
i
)|
2
→ min
0≤α
k
(t
i
)≤1
, k = 1, 2, . . . , n i = 1, 2, . . . , N.
The Case of Single Alpha (α)
In the case of single absorption, consideration of particular models leads to the following ex-
pression for Π
BTC
:
Π
BTC
=
∂
∂t
"
ln
n
X
k=1
α · P
k
· R
k
!
+ ln
1
n
n
X
j=1
T
′
j
!
− ln b − ln h + ln d
#
=
=
∂
∂t
"
ln
α ·
n
X
k=1
P
k
· R
k
!
+ ln
1
n
n
X
j=1
T
′
j
!
− ln b − ln h + ln d
#
=
=
∂
∂t
"
ln α + ln
n
X
k=1
P
k
· R
k
!
+ ln
1
n
n
X
j=1
T
′
j
!
− ln b − ln h + ln d
#
.
Similarly, in this case, we have
ln P
BTC
= ln α + ln
n
X
k=1
P
k
· R
k
!
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d.
Numerical Analysis
In this section, we are going to consider the determination of single and individual absorption
rates. To this end, we are going to consider a specific database of 11 markets, i.e., n = 11.
Based on the database structure, the time variable has the following form:
t
i
= 7 +
i
365
,
with N = 1592 data points. In this particular case, m = 1, so that there is only T
′
1
.
In this case, we have a system of 1592 equations with respect to 11·1592 = 17512 unknowns,
which are computed by minimizing the l
1
-norm of the residue:
Err (t
i
) = |F (α
1
(t
i
) , α
2
(t
i
) , . . . , α
n
(t
i
) ; t
i
) − S (t
i
)| → min
0≤α
k
(t
i
)≤1
,
with k = 1, 2, . . . , 11, and i = 1, 2, . . . , 1592.
The numerical values of α
k
(t
i
) are plotted in Figures 1 and 2 below. The corresponding
error function Err is plotted in Figure 11, as well.
7 8 9 10 11
0.0
0.2
0.4
0.6
0.8
1.0
t
α
7
× 10
14
7 8 9 10 11
0.0
0.2
0.4
0.6
0.8
1.0
t
α
8
× 10
14
7 8 9 10 11
0.0
0.2
0.4
0.6
0.8
1.0
t
α
9
× 10
14
7 8 9 10 11
0.0
0.2
0.4
0.6
0.8
1.0
t
α
10
× 10
14
7 8 9 10 11
0.0
0.2
0.4
0.6
0.8
1.0
t
α
11
× 10
14
7 8 9 10 11
0.00
0.02
0.04
0.06
0.08
0.10
t
Err
Figure 2: Dependence of α
7
, . . ., α
11
and Err on t according to (12)
Thus, the numerical analysis in the above particular case shows similar behavior and range
of values for α
1
, α
2
, . . ., α
11
making the dependence of the absorption rate on a specific market
very weak. This apparently suggests that, in general, for n markets, we can make an assumption
that
α
1
≈ α
2
≈ . . . ≈ α
n
:= α,
which will simplify the rigorous analysis significantly.
Our aim now is to show that the case of individual absorption rates can be, to some extent,
approximated by a single absorption rate for all markets. In other words, we are going to
assume that the absorption rate introduced in (9) does not depend on the subscript k. Then,
n
X
k=1
α
k
· P
k
· R
k
= α ·
n
X
k=1
P
k
· R
k
,
where P
k
and R
k
are given, α
k
are determined in the previous section and α is unknown
determined as follows:
α =
n
X
k=1
α
k
· P
k
· R
k
n
X
k=1
P
k
· R
k
. (13)
Figure 3 below shows the dependence of α and ln α on t.
7 8 9 10 11
0.0
0.2
0.4
0.6
0.8
1.0
t
α × 10
14
7 8 9 10 11
-36
-35
-34
-33
-32
-31
t
ln α
Figure 3: Dependence of α and ln α on t according to (13)
In order to estimate the error of approximation by single absorption rate, we measure the
error
MSE =
1
2
α
n
X
k=1
P
k
· R
k
− S
!
2
. (14)
Sensitivity Analysis - General Case
In this section, we carry out sensitivity analysis for Π
BTC
with respect to all parameters included
in it. We start with the general case and the incorporate some specific models and assumptions.
The Case of the Generalised Model
First, consider the general case when
Π
BTC
=
∂
∂t
"
ln
n
X
k=1
P
k
· U
s
k
!
+ ln V
BTC
− ln Q
BTC
#
.
Now, before proceeding to the sensitivity analysis, we assume that there is no correlation
between f
′
and f for none of the functions entering the expression of Π
BTC
.
Detailed calculation of derivatives used for the sensitivity analysis carried out in this section
are presented at the end of this Appendix.
Sensitivity with Respect to Velocity
In this section, we study the sensitivity of Π
BTC
with respect to V
BTC
and V
′
BTC
. First note
that since
∂Π
BTC
∂V
BTC
= −
V
′
BTC
V
2
BTC
.
Taking into account that the denominator is always positive, we come to the evident conclusion
that
∂Π
BTC
∂V
BTC
> 0 when V
′
BTC
< 0
and
∂Π
BTC
∂V
BTC
< 0 when V
′
BTC
> 0.
In other words, when V
BTC
is a decreasing (increasing) function of t, then Π
BTC
is an increasing
(decreasing) function of V
BTC
.
On the other hand, since
∂Π
BTC
∂V
′
BTC
=
1
V
BTC
which is apparently always positive, we conclude that Π
BTC
is an increasing (linear) function
of V
′
BTC
for all values of t.
Sensitivity with Respect to Output
According to the derivative expression derived at the end of this section,
∂Π
BTC
∂Q
BTC
=
Q
′
BTC
Q
2
BTC
.
Taking into account that Q
BTC
is an increasing function of t, implying that Q
′
BTC
> 0, we
conclude that
∂Π
BTC
∂Q
BTC
> 0
for all t. In other words, Π
BTC
is always an increasing function of Q
BTC
.
Similarly, since
∂Π
BTC
∂Q
′
BTC
= −
1
Q
BTC
,
we see that
∂Π
BTC
∂Q
′
BTC
< 0
for all t. Therefore, Π
BTC
is a decreasing function of Q
′
BTC
.
Sensitivity with Respect to Asset Prices
Taking into account that
∂Π
BTC
∂P
′
k
=
U
s
k
S
,
in which
S =
n
X
k=1
P
k
· U
s
k
,
it becomes evident that
∂Π
BTC
∂P
′
k
> 0
for all t. Thence, Π
BTC
is an increasing function of P
′
k
.
On the other hand, since
∂Π
BTC
∂P
k
=
U
s
k
S
·
∂
∂t
ln
U
s
k
S
,
we observe that when
U
s
k
S
is an increasing function of t, then
∂
∂t
ln
U
s
k
S
> 0
implying that
∂Π
BTC
∂P
k
> 0.
Similarly, when
U
s
k
S
is a decreasing function of t, then
∂Π
BTC
∂P
k
< 0.
Thence, when
U
s
k
S
is an increasing (decreasing) function of t, Π
BTC
is an increasing (decreasing)
function of P
k
.
Sensitivity with Respect to Absorption (U
s
k
)
Since
∂Π
BTC
∂U
′
s
k
=
P
k
S
> 0
for all t. Therefore, Π
BTC
is an increasing function of U
′
s
k
.
On the other hand,
∂Π
BTC
∂U
s
k
=
P
k
S
·
∂
∂t
ln
P
k
S
providing that when
P
k
S
is an increasing (decreasing) function of t, Π
BTC
is an increasing
(decreasing) function of U
s
k
.
Sensitivity Analysis - Market Specific Alpha Case
In this section, we are going to study the sensitivity of Π
BTC
with respect to parameters
introduced when considering the particular models above. To this end, we are going to make
use of the following expression for Π
BTC
:
Π
BTC
=
∂
∂t
"
ln
n
X
k=1
α
k
· P
k
· R
k
!
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d
#
.
Sensitivity Analysis with Respect to Volume
Since
∂Π
BTC
∂R
k
=
W
k
R
·
∂
∂t
ln
W
k
R
,
where
W
k
= P
k
· α
k
, R =
n
X
k=1
P
k
· R
k
,
then, we straightforwardly conclude that when
W
k
R
is an increasing (decreasing) function of t,
then Π
BTC
is an increasing (decreasing) function of R
k
.
On the other hand,
∂Π
BTC
∂R
′
k
=
W
k
R
,
leading to the conclusion that Π
BTC
is an increasing function of R
′
k
.
Sensitivity with Respect to Velocity
Taking into account that
∂Π
BTC
∂T
′
j
= −
1
T
′
j
2
·
m
X
l=1
T
′′
l
,
and taking into account that the denominator is positive for all t, then it is obvious that
n
X
m=1
T
′′
m
< 0
implies that
∂Π
BTC
∂T
′
j
> 0,
i.e., Π
BTC
is an increasing function of T
′
j
. On the other hand,
n
X
m=1
T
′′
m
> 0
implies that
∂Π
BTC
∂T
′
j
< 0,
i.e., Π
BTC
is a decreasing function of T
′
j
.
Sensitivity with Respect to Output Parameters
Taking into account that
∂Π
BTC
∂b
=
b
′
b
2
,
the denominator of which is positive for all t, we conclude that
∂Π
BTC
∂b
> 0
when b
′
> 0. Similarly,
∂Π
BTC
∂b
< 0
when b
′
< 0.
Hence, when b is an increasing (decreasing) function of t, then Π
BTC
is an increasing (de-
creasing) function of b.
Using the expression
∂Π
BTC
∂h
=
h
′
h
2
,
we conclude that, when h is an increasing (decreasing) function of t, then Π
BTC
is an increasing
(decreasing) function of h.
However, since
∂Π
BTC
∂d
= −
d
′
d
2
,
we come to the conclusion that when d is an increasing (decreasing) function of t, then Π
BTC
is a decreasing (increasing) function of d.
Sensitivity Analysis - Single Alpha Case
In this section, we consider the sensitivity analysis in the single α , i.e., we assume that the
absorption rate is the same for all markets. Here we will use actual raw market data to determine
relational effects.
In the case of single absorption, consideration of particular models leads to the following
expression for Π
BTC
:
Π
BTC
=
∂
∂t
"
ln α + ln
n
X
k=1
P
k
· R
k
!
+ ln
1
n
n
X
j=1
T
′
j
!
− ln b − ln h + ln d
#
.
Therefore, we have
ln P
BTC
= ln α + ln
n
X
k=1
P
k
· R
k
!
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d.
Sensitivity of ln P
BTC
Using the simple formulas obtained in Appendix 1, we come to the following conclusions.
1. ln P
BTC
is an increasing function of α for all values of t. See Figure 5 (a).
2. ln P
BTC
is an increasing function of P
k
and R
k
for all values of k and t.
3. ln P
BTC
is always an increasing function of P . See Figure 5 (b).
4. ln P
BTC
is an increasing function of T
′
j
for all values of j and t. See Figure 5 (c).
5. ln P
BTC
is a decreasing function of b and h, and it is an increasing function of d for all
values of t. See Figure 5 (d-f).
0.0 0.5 1.0 1.5 2.0
7
8
9
10
11
α 10
14
ln P
BTC
(a)
0.0 0.5 1.0 1.5 2.0
7
8
9
10
11
P 10
-13
ln P
BTC
(b)
0 1 2 3 4 5 6
7
8
9
10
11
T
1
10
-5
ln P
BTC
(c)
6 7 8 9 10 11 12
7
8
9
10
11
b
ln P
BTC
(d)
0.0 0.5 1.0 1.5 2.0
7
8
9
10
11
h 10
-8
ln P
BTC
(e)
0.0 0.5 1.0 1.5 2.0
7
8
9
10
11
d 10
-13
ln P
BTC
(f)
Figure 5: Sensitivity plots of ln P
BTC
with respect to system parameters
0.6 0.7 0.8 0.9
7
8
9
10
11
k =1
n
P
k
10
-5
ln P
BTC
(a)
0.0 0.5 1.0 1.5 2.0 2.5
7
8
9
10
11
k =1
n
R
k
10
-9
ln P
BTC
(b)
Figure 6: Sensitivity plots of ln P
BTC
with respect to sum of P
k
(a) and R
k
(b)
Sensitivity of Π
BTC
Sensitivity with Respect to Absorption Rate
In this case,
∂Π
BTC
∂α
= −
α
′
α
2
,
leading us to the conclusion that when α is an increasing function of t, i.e., when α
′
> 0, Π
BTC
is a decreasing function of α. And, when α is a decreasing function of t, i.e., when α
′
< 0,
Π
BTC
is an increasing function of α.
On the other hand,
∂Π
BTC
∂α
′
=
1
α
.
Since α > 0 for all t, we conclude that Π
BTC
is an increasing function of α
′
for all values of t.
Sensitivity with Respect to Price and Asset Volume
Due to symmetry of Π
BTC
with respect to P
k
and R
k
, we derive similar formulas for
∂Π
BTC
∂P
k
=
R
k
P
[1 − P
′
]
and
∂Π
BTC
∂R
k
=
P
k
P
[1 − P
′
]
,
where
P =
n
X
m=1
P
m
· R
m
.
Apparently, the sign of both
∂Π
BTC
∂P
k
and
∂Π
BTC
∂R
k
depends on the sign of the expression
1 − P
′
. More specifically, when 1 − P
′
> 0, we get
∂Π
BTC
∂P
k
> 0 and
∂Π
BTC
∂R
k
> 0,
meaning that Π
BTC
is an increasing function of P
k
and R
k
simultaneously. On the other hand,
1 − P
′
< 0 leads to
∂Π
BTC
∂P
k
< 0 and
∂Π
BTC
∂R
k
< 0,
meaning that Π
BTC
is a decreasing function of P
k
and R
k
simultaneously.
Note that Π
BTC
is symmetric also with respect to P
′
k
and R
′
k
with
∂Π
BTC
∂P
′
k
=
R
k
P
and
∂Π
BTC
∂R
′
k
=
P
k
P
.
Apparently, both expressions are positive for all values of t leading to the conclusion that Π
BTC
is an increasing function of both P
′
k
and R
′
k
.
Sensitivity with Respect to Sum
On the basis of the expression
∂Π
BTC
∂P
= −
P
′
P
2
,
we conclude that when P is an increasing (decreasing) function of time, i.e., P
′
> 0 (P
′
< 0,
respectively), then Π
BTC
is a decreasing (increasing) function of P .
Sensitivity with Respect to Transactions
The sensitivity of Π
BTC
with respect to transactions is more complicated to explore, since the
corresponding derivative is given by
∂Π
BTC
∂T
′
j
= −
1
T
′
j
2
·
m
X
l=1
T
′′
l
.
Therefore, taking into account that
T
′
j
2
is positive for all values of t, we come to the conclusion
that when
m
X
l=1
T
′′
l
< 0,
then Π
BTC
is an increasing function of T
′
j
, and when
m
X
l=1
T
′′
l
> 0,
evidently, Π
BTC
is a decreasing function of T
′
j
.
Sensitivity with Respect to Output Parameters
In Appendix 1 we derive that
∂Π
BTC
∂b
=
b
′
b
2
,
∂Π
BTC
∂h
=
h
′
h
2
and
∂Π
BTC
∂d
= −
d
′
d
2
.
And for the derivatives,
∂Π
BTC
∂b
′
= −
1
b
,
∂Π
BTC
∂h
′
= −
1
h
and
∂Π
BTC
∂d
′
=
1
d
.
Note that b > 0, h > 0 and d > 0 for all values of t.
Therefore, we straightforwardly conclude:
1. When b
′
> 0 (b
′
< 0), i.e., b is an increasing (decreasing) function of t, then Π
BTC
is an
increasing (decreasing) function of b.
2. When h
′
> 0 (h
′
< 0), i.e., h is an increasing (decreasing) function of t, then Π
BTC
is an
increasing (decreasing) function of h.
3. When d
′
> 0 (d
′
< 0), i.e., d is a decreasing (increasing) function of t, then Π
BTC
is a
decreasing (increasing) function of d.
4. Π
BTC
is a decreasing function of b
′
and h
′
and it is an increasing function of d
′
for all
values of t.
Numerical Analysis for Particular Data
In this section, we are going to verify the conclusions obtained in the previous section on
a particular database. To this aim, in Figure 7, we plot the dependence of corresponding
derivative of Π
BTC
with respect to sensitivity parameter.
Specifically, Figure 7 (a) shows that when α increases, α
′
becomes negative. Therefore,
since
∂Π
BTC
∂α
= −
α
′
α
2
,
we conclude that for large α, Π
BTC
is an increasing function of α.
An important conclusion is based on Figure 7 (b). Evidently, 1 − P
′
becomes positive when
P increases. Therefore, taking into account that
∂Π
BTC
∂P
k
=
R
k
P
· (1 − P
′
) ,
∂Π
BTC
∂R
k
=
P
k
P
· (1 − P
′
) ,
we obtain that Π
BTC
increases with respect to P
k
and R
k
for all k when P increases.
Similar behavior is observed for P as shown on Figure 7 (c). Since P
′
becomes negative for
large values of P , and taking into account that
∂Π
BTC
∂P
= −
P
′
P
2
,
we conclude that for large values of P , Π
BTC
is an increasing function of P .
Figure 7 (c) shows similar behavior for T
′
1
leading to the conclusion that for large values of
T
′
1
, Π
BTC
is an increasing function of T
′
1
.
On the other hand, since
∂Π
BTC
∂d
= −
d
′
d
2
,
and as Figure 7 (e) shows, d
′
is positive for large values of d, we come to the conclusion that
for large values of d, Π
BTC
is an increasing function of d.
0.0 0.1 0.2 0.3 0.4
-0.2
-0.1
0.0
0.1
0.2
α 10
13
α' 10
10
(a)
0.0 0.5 1.0 1.5
-0.2
0.0
0.2
0.4
P 10
-13
(1 - P') 10
-16
(b)
0.0 0.5 1.0 1.5
-0.4
-0.2
0.0
0.2
P 10
-13
P' 10
-16
(c)
0.0 0.2 0.4 0.6 0.8
-1.0
-0.5
0.0
0.5
1.0
T
1
10
-13
T
1
''
10
-16
(d)
0.0 0.5 1.0 1.5 2.0
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
d 10
-13
d' 10
-15
(e)
0.0 0.5 1.0 1.5 2.0
-1.0
-0.5
0.0
0.5
1.0
h 10
-8
h' 10
-10
(f)
Figure 7: Sensitivity plots of α
′
(a), 1 − P
′
(b), P
′
(c), T
′
1
(d), d
′
(e) and h
′
(f)
Derivations
We collect the derivative calculations that are used in sensitivity analysis with respect to system
parameters.
Derivative of ln P
BTC
We now proceed with evaluation of derivatives of the expression
ln P
BTC
= ln α + ln
n
X
k=1
P
k
· R
k
!
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d.
Apparently, ln P
BTC
does not depend on parameter derivatives explicitly.
With Respect to α
We easily compute
∂ ln P
BTC
∂α
=
∂ ln α
∂α
=
1
α
.
With Respect to Price and Volume
It is easy to see that ln P
BTC
is symmetric with respect to P
k
and R
k
, leading to
∂ ln P
BTC
∂P
k
=
1
P
∂
∂P
k
(P
k
· R
k
) =
R
k
P
and
∂ ln P
BTC
∂R
k
=
1
P
∂
∂R
k
(P
k
· R
k
) =
P
k
P
.
With Respect to Sum
In this case,
∂ ln P
BTC
∂P
= 1.
With Respect to Transactions
Due to linear dependence of ln P
BTC
on T
′
j
, we obtain
∂ ln P
BTC
∂T
′
j
=
1
m
∂
∂T
′
j
m
X
j=1
T
′
j
!
=
1
m
.
With Respect to Output Parameters
In the case of output parameters, we easily derive
∂ ln P
BTC
∂b
= −
∂ ln b
∂b
= −
1
b
,
∂ ln P
BTC
∂h
= −
∂ ln h
∂h
= −
1
h
,
∂ ln P
BTC
∂d
=
∂ ln d
∂d
=
1
d
.
Derivative of Π
BTC
We consider the expression
Π
BTC
=
∂
∂t
"
ln α + ln
n
X
k=1
P
k
· R
k
!
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d
#
.
With Respect to α
Apparently, only the first term in the expression for Π
BTC
depends on α explicitly. Therefore,
we get
∂Π
BTC
∂α
=
∂
∂α
α
′
α
= −
α
′
α
2
.
On the other hand,
∂Π
BTC
∂α
′
=
∂
∂α
′
α
′
α
=
1
α
.
With Respect to Price
In this case,
∂Π
BTC
∂P
k
=
∂
∂P
k
n
X
k=1
(P
k
· R
k
)
′
n
X
k=1
P
k
R
k
=
∂
∂P
k
n
X
k=1
(P
′
k
· R
k
+ P
k
· R
′
k
)
n
X
k=1
P
k
R
k
=
= −
1
P
2
·
∂P
∂P
k
·
n
X
k=1
(P
′
k
· R
k
+ P
k
· R
′
k
) +
1
P
∂
∂P
k
"
n
X
k=1
(P
′
k
· R
k
+ P
k
· R
′
k
)
#
=
= −
1
P
2
· R
k
·
n
X
m=1
(P
′
m
· R
m
+ P
m
· R
′
m
) +
1
P
· R
k
=
=
R
k
P
"
−
n
X
m=1
(P
′
m
· R
m
+ P
m
· R
′
m
) + 1
#
=
R
k
P
[1 − P
′
] .
Here,
P =
n
X
k=1
P
k
· R
k
.
On the other hand, Π
BTC
is a linear function with respect to P
′
k
with the linear coefficient
∂Π
BTC
∂P
′
k
=
∂
∂P
′
k
"
1
P
n
X
k=1
(P
′
k
· R
k
+ P
k
· R
′
k
)
#
=
R
k
P
.
With Respect to Volume
And in a similar way we compute
∂Π
BTC
∂R
k
=
∂
∂R
k
n
X
k=1
(P
k
· R
k
)
′
n
X
k=1
P
k
R
k
=
∂
∂R
k
n
X
k=1
(P
′
k
· R
k
+ P
k
· R
′
k
)
n
X
k=1
P
k
R
k
=
= −
1
P
2
·
∂P
∂R
k
·
n
X
k=1
(P
′
k
· R
k
+ P
k
· R
′
k
) +
1
P
∂
∂R
k
"
n
X
k=1
(P
′
k
· R
k
+ P
k
· R
′
k
)
#
=
= −
1
P
2
· P
k
·
n
X
m=1
(P
′
m
· R
m
+ P
m
· R
′
m
) +
1
P
· P
k
=
=
P
k
P
"
−
n
X
m=1
(P
′
m
· R
m
+ P
m
· R
′
m
) + 1
#
=
P
k
P
[1 − P
′
] .
Similar to the case of P
′
k
, Π
BTC
is a linear function with respect to R
′
k
as well. At this, the
linear coefficient is defined as follows:
∂Π
BTC
∂R
′
k
=
∂
∂R
′
k
"
1
P
n
X
k=1
(P
′
k
· R
k
+ P
k
· R
′
k
)
#
=
P
k
P
.
With Respect to Sum
The case of derivative with respect to P is even simpler. Indeed,
∂Π
BTC
∂P
=
∂
∂P
P
′
P
= −
P
′
P
2
.
With Respect to Transactions
In this case,
∂Π
BTC
∂T
′
j
=
∂
∂T
′
j
"
∂
∂t
"
ln
1
m
m
X
j=1
T
′
j
!##
=
∂
∂T
′
j
m
X
j=1
T
′′
j
m
X
j=1
T
′
j
= −
1
T
′
j
2
·
m
X
l=1
T
′′
l
.
With Respect to Output Parameters
We easily evaluate the following partial derivatives:
∂Π
BTC
∂b
= −
∂
∂b
∂ ln b
∂t
= −
∂
∂b
b
′
b
=
b
′
b
2
,
∂Π
BTC
∂h
= −
∂
∂h
∂ ln h
∂t
=
h
′
h
2
,
and
∂Π
BTC
∂d
=
∂
∂d
∂ ln d
∂t
= −
d
′
d
2
.
And for the derivatives we have
∂Π
BTC
∂b
′
= −
∂
∂b
′
b
′
b
= −
1
b
,
∂Π
BTC
∂h
′
= −
∂
∂h
′
h
′
h
= −
1
h
,
and
∂Π
BTC
∂d
′
=
∂
∂d
′
d
′
d
=
1
d
.
Further Derivations
With Respect to Velocity
Here we present derivations of some partial derivatives required for sensitivity analysis carried
out in previous sections.
In order to study the sensitivity of Π
BTC
with respect to V
BTC
, we compute
∂Π
BTC
∂V
BTC
=
∂
∂V
BTC
"
∂ ln V
BTC
∂t
+
∂
∂t
"
ln
n
X
k=1
P
k
· U
s
k
!
− ln Q
BTC
##
=
=
∂
∂V
BTC
∂ ln V
BTC
∂t
=
∂
∂V
BTC
V
′
BTC
V
BTC
= −
V
′
BTC
V
2
BTC
.
On the other hand, the sensitivity of Π
BTC
with respect to V
′
BTC
is carried out on the basis of
the derivative
∂Π
BTC
∂V
′
BTC
=
∂
∂V
′
BTC
"
∂ ln V
BTC
∂t
+
∂
∂t
"
ln
n
X
k=1
P
k
· U
s
k
!
− ln Q
BTC
##
=
=
∂
∂V
′
BTC
∂ ln V
BTC
∂t
=
∂
∂V
′
BTC
V
′
BTC
V
BTC
=
1
V
BTC
.
With Respect to Output
In this case, we have
∂Π
BTC
∂Q
BTC
=
∂
∂Q
BTC
"
−
∂ ln Q
BTC
∂t
+
∂
∂t
"
ln
n
X
k=1
P
k
· U
s
k
!
+ ln V
BTC
##
=
= −
∂
∂Q
BTC
∂ ln Q
BTC
∂t
= −
∂
∂Q
BTC
Q
′
BTC
Q
BTC
=
Q
′
BTC
Q
2
BTC
.
In the same way, we compute
∂Π
BTC
∂Q
′
BTC
=
∂
∂Q
BTC
"
−
∂ ln Q
BTC
∂t
+
∂
∂t
"
ln
n
X
k=1
P
k
· U
s
k
!
+ ln V
BTC
##
=
= −
∂
∂Q
′
BTC
∂ ln Q
BTC
∂t
= −
∂
∂Q
′
BTC
Q
′
BTC
Q
BTC
= −
1
Q
BTC
.
With Respect to Asset Price
We start with simplifying the corresponding term in the expression of Π
BTC
,
∂
∂t
"
ln
n
X
k=1
P
k
· U
s
k
!#
=
n
X
k=1
(P
k
· U
s
k
)
′
n
X
k=1
P
k
· U
s
k
=
1
S
·
n
X
k=1
P
′
k
· U
s
k
+ P
k
· U
′
s
k
=
=
1
S
·
n
X
k=1
P
′
k
· U
s
k
+
1
S
·
n
X
k=1
P
k
· U
′
s
k
,
in which
S =
n
X
k=1
P
k
· U
s
k
.
As a matter of fact, Π
BTC
is linear in P
′
k
. Therefore, the derivative of Π
BTC
with respect to
that variable is easy to compute. Indeed,
∂Π
BTC
∂P
′
k
=
∂
∂P
′
k
"
∂
∂t
"
ln
n
X
k=1
P
k
· U
s
k
!#
+
∂
∂t
[ln V
BTC
− ln Q
BTC
]
#
=
=
∂
∂P
′
k
"
1
S
·
n
X
k=1
P
′
k
· U
s
k
+
1
S
·
n
X
k=1
P
k
· U
′
s
k
#
=
U
s
k
S
.
On the other hand,
∂Π
BTC
∂P
k
=
∂
∂P
k
"
∂
∂t
"
ln
n
X
k=1
P
k
· U
s
k
!#
+
∂
∂t
[ln V
BTC
− ln Q
BTC
]
#
=
=
∂
∂P
k
"
1
S
·
n
X
k=1
P
′
k
· U
s
k
+
1
S
·
n
X
k=1
P
k
· U
′
s
k
#
=
=
∂
∂P
k
1
S
·
n
X
k=1
P
′
k
· U
s
k
+
1
S
·
∂
∂P
k
"
n
X
k=1
P
′
k
· U
s
k
#
+
+
∂
∂P
k
1
S
·
n
X
k=1
P
k
· U
′
s
k
+
1
S
·
∂
∂P
k
"
n
X
k=1
P
k
· U
′
s
k
#
.
Let us compute the derivatives above one by one:
∂
∂P
k
1
S
= −
1
S
2
·
∂
∂P
k
"
n
X
k=1
P
k
· U
s
k
#
= −
U
s
k
S
2
,
∂
∂P
k
"
n
X
k=1
P
′
k
· U
s
k
#
= 0,
∂
∂M
k
"
n
X
k=1
P
k
· U
′
s
k
#
= U
′
s
k
.
Substituting these expressions into
∂Π
BTC
∂P
k
, we obtain
∂Π
BTC
∂P
k
=
∂
∂P
k
1
S
·
n
X
k=1
P
′
k
· U
s
k
+
1
S
·
∂
∂P
k
"
n
X
k=1
P
′
k
· U
s
k
#
+
∂
∂P
k
1
S
·
n
X
k=1
P
k
· U
′
s
k
=
= −
U
s
k
S
2
·
n
X
l=1
P
′
l
· U
s
l
+
U
′
s
k
S
−
U
s
k
S
2
·
n
X
l=1
P
l
· U
′
s
l
=
=
U
s
k
S
·
"
U
′
s
k
U
s
k
· S −
∂
∂t
n
X
l=1
P
l
· U
s
l
!#
=
=
U
s
k
S
2
· S ·
U
′
s
k
U
s
k
−
∂ ln S
∂t
=
=
U
s
k
S
·
∂
∂t
[ln U
s
k
− ln S]
or
∂Π
BTC
∂P
k
=
U
s
k
S
·
∂
∂t
ln
U
s
k
S
.
With Respect to Absorption
In this case as well, we are going to use the expression
Π
BTC
=
1
S
·
n
X
k=1
P
′
k
· U
s
k
+
1
S
·
n
X
k=1
P
k
· U
′
s
k
+
∂
∂t
[ln V
BTC
− ln Q
BTC
] .
Then, obviously,
∂Π
BTC
∂U
′
s
k
=
∂
∂U
′
s
k
"
1
S
·
n
X
k=1
P
′
k
· U
s
k
+
1
S
·
n
X
k=1
P
k
· U
′
s
k
#
=
P
k
S
.
On the other hand,
∂Π
BTC
∂U
s
k
=
∂
∂U
s
k
"
1
S
·
n
X
k=1
P
′
k
· U
s
k
+
1
S
·
n
X
k=1
P
k
· U
′
s
k
+
∂
∂t
[ln V
BTC
− ln Q
BTC
]
#
=
=
∂
∂U
s
k
1
S
·
n
X
k=1
P
′
k
· U
s
k
+
1
S
·
∂
∂U
s
k
"
n
X
k=1
P
′
k
· U
s
k
#
+
+
∂
∂U
s
k
1
S
·
n
X
k=1
P
k
· U
′
s
k
+
1
S
·
∂
∂U
s
k
"
n
X
k=1
P
k
· U
′
s
k
#
.
Apparently,
∂
∂U
s
k
1
S
= −
P
k
S
2
,
∂
∂U
s
k
"
n
X
k=1
P
′
k
· U
s
k
#
= P
′
k
,
∂
∂U
s
k
"
n
X
k=1
P
k
· U
′
s
k
#
= 0.
Therefore,
∂Π
BTC
∂U
s
k
= −
P
k
S
2
·
n
X
k=1
P
′
k
· U
s
k
−
P
k
S
2
·
n
X
k=1
P
k
· U
′
s
k
+
P
′
k
S
=
=
P
k
S
·
"
−
1
S
·
n
X
k=1
P
′
k
· U
s
k
−
1
S
·
n
X
k=1
P
k
· U
′
s
k
+
P
′
k
P
k
#
=
=
P
k
S
·
"
−
1
S
·
n
X
k=1
P
′
k
· U
s
k
+
n
X
k=1
P
k
· U
′
s
k
!
+
P
′
k
P
k
#
=
=
P
k
S
·
"
P
′
k
P
k
−
1
S
·
∂
∂t
n
X
k=1
P
k
· U
s
k
!#
=
=
P
k
S
·
P
′
k
P
k
−
P
′
k
S
=
P
k
S
·
∂
∂t
[ln P
k
− ln S]
or
∂Π
BTC
∂U
s
k
=
P
k
S
·
∂
∂t
ln
P
k
S
.
With Respect to Volume
In this case,
∂Π
BTC
∂R
k
=
∂
∂R
k
"
∂
∂t
ln
n
X
k=1
P
k
· R
k
!#
=
∂
∂R
k
n
X
k=1
W
k
· R
k
!
′
n
X
k=1
W
k
· R
k
=
=
∂
∂R
k
"
1
R
·
n
X
k=1
(W
′
k
· R
k
+ W
k
· R
′
k
)
#
,
where
W
k
= P
k
· α
k
, R =
n
X
k=1
P
k
· R
k
.
Therefore,
∂Π
BTC
∂R
k
=
∂
∂R
k
"
1
R
·
n
X
k=1
(W
′
k
· R
k
+ W
k
· R
′
k
)
#
=
=
∂
∂R
k
1
R
·
n
X
k=1
(W
′
k
· R
k
+ W
k
· R
′
k
) +
1
R
·
∂
∂R
k
"
n
X
k=1
(W
′
k
· R
k
+ W
k
· R
′
k
)
#
.
Taking into account the chain rule, we obtain
∂
∂R
k
1
R
= −
1
R
2
·
∂R
∂R
k
= −
W
k
R
2
.
On the other hand,
∂
∂R
k
"
n
X
k=1
(W
′
k
· R
k
+ W
k
· R
′
k
)
#
= W
′
k
,
providing us with this final form:
∂Π
BTC
∂R
k
= −
W
k
R
2
·
n
X
k=1
(W
′
k
· R
k
+ W
k
· R
′
k
) +
P
′
k
R
=
=
W
k
R
·
"
−
1
R
·
n
X
k=1
(W
′
k
· R
k
+ W
k
· R
′
k
) +
W
′
k
W
k
#
=
=
W
k
R
·
−
R
′
R
+
W
′
k
W
k
=
=
W
k
R
·
∂
∂t
[ln W
k
− ln R]
or
∂Π
BTC
∂R
k
=
W
k
R
·
∂
∂t
ln
W
k
R
.
It is much easier to establish that
∂Π
BTC
∂R
′
k
=
∂
∂R
′
k
"
1
R
·
n
X
k=1
(W
′
k
· R
k
+ W
k
· R
′
k
)
#
=
W
k
R
,
Bass Model for Alpha
For the absorption rate α
k
we consider the following model based on the well-known Bass
Diffusion
α
k
=
1 − exp [− (p
k
+ q
k
) t]
1 +
p
k
q
k
· exp [− (p
k
+ q
k
) t]
,
where p
k
and q
k
are time-dependent coefficients to be estimated.
Then, Π
BTC
will obtain the following form:
Π
BTC
=
∂
∂t
ln
n
X
k=1
P
k
·
1 − exp [− (p
k
+ q
k
) t]
1 +
p
k
q
k
· exp [− (p
k
+ q
k
) t]
· R
k
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d
.
Bass Derivatives
It is easy to see that
∂Π
BTC
∂p
k
=
∂
∂p
k
X
′
X
=
1
X
·
∂X
′
∂p
k
+ X
′
·
∂
∂p
k
1
X
=
1
X
·
∂X
′
∂p
k
−
X
′
X
2
·
∂X
∂p
k
,
and, similarly,
∂Π
BTC
∂q
k
=
1
X
·
∂X
′
∂q
k
−
X
′
X
2
·
∂X
∂q
k
,
where
X =
n
X
k=1
P
k
· R
k
· q
k
·
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
.
Therefore,
X
′
=
∂X
∂t
=
n
X
k=1
(P
k
· R
k
)
′
· q
k
·
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
+
+
n
X
k=1
P
k
· R
k
· q
′
k
·
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
+
+
n
X
k=1
P
k
· R
k
· q
k
·
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
′
.
Since p
k
and q
k
depend on t, we compute
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
′
=
(1 − exp [− (p
k
+ q
k
) t])
′
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
−
−
(1 − exp [− (p
k
+ q
k
) t]) (q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
′
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
.
Now, making use of the chain rule, it is easy to evaluate
(1 − exp [− (p
k
+ q
k
) t])
′
= − exp [− (p
k
+ q
k
) t] · [− (p
k
+ q
k
) t]
′
=
= [p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t] · exp [− (p
k
+ q
k
) t] ,
and
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
′
= q
′
k
+ p
′
k
· exp [− (p
k
+ q
k
) t] +
+ p
k
· exp [− (p
k
+ q
k
) t] · [− (p
k
+ q
k
) · t]
′
=
= q
′
k
+ p
′
k
· exp [− (p
k
+ q
k
) t] +
− p
k
· exp [− (p
k
+ q
k
) t] · [p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t] =
= q
′
k
+ [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t]) · exp [− (p
k
+ q
k
) t] .
Hence,
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
′
=
(p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
−
−
q
′
k
+ [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t]) · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
.
Thus,
X
′
=
∂X
∂t
=
n
X
k=1
(P
k
· R
k
)
′
· q
k
·
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
+
+
n
X
k=1
P
k
· R
k
· q
′
k
·
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
+
+
n
X
k=1
P
k
· R
k
· q
k
·
(p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
−
−
n
X
k=1
P
k
· R
k
· q
k
·
q
′
k
+ [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)] · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
.
Now, in order to continue the sensitivity analysis of Π
BTC
with respect to p
k
, we need to
evaluate the derivatives
∂X
∂p
k
and
∂X
′
∂p
k
. First,
∂X
∂p
k
=
∂
∂p
k
P
k
· R
k
· q
k
·
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
=
= P
k
· R
k
· q
k
·
∂
∂p
k
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
,
in which
∂
∂p
k
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
=
1
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
∂
∂p
k
[1 − exp [− (p
k
+ q
k
) t]] +
+ (1 − exp [− (p
k
+ q
k
) t]) ·
∂
∂p
k
1
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
=
= −
exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
·
∂
∂p
k
[− (p
k
+ q
k
) t] −
−
1 − exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
·
∂
∂p
k
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t]) =
=
t · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
−
−
1 − exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
·
exp [− (p
k
+ q
k
) t] + p
k
·
∂ exp [− (p
k
+ q
k
) t]
∂p
k
=
=
t · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
−
−
(1 − exp [− (p
k
+ q
k
) t]) · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
·
1 + p
k
·
∂
∂p
k
[− (p
k
+ q
k
) t]
=
=
t · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
−
(1 − exp [− (p
k
+ q
k
) t]) · exp [− (p
k
+ q
k
) t] · (1 − p
k
· t)
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
=
=
exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
t −
(1 − exp [− (p
k
+ q
k
) t]) · (1 − p
k
· t)
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
=
=
exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
×
× [t · q
k
+ t · p
k
· exp [− (p
k
+ q
k
) t] − (1 − exp [− (p
k
+ q
k
) t]) · (1 − p
k
· t)] =
=
exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
×
× [t · q
k
+ t · p
k
· exp [− (p
k
+ q
k
) t] − (1 − p
k
· t − exp [− (p
k
+ q
k
) t] + p
k
· t · exp [− (p
k
+ q
k
) t])] =
=
exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
· [(p
k
+ q
k
) · t + exp [− (p
k
+ q
k
) t] − 1] ,
or
∂
∂p
k
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
=
exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
×
× [(p
k
+ q
k
) · t + exp [− (p
k
+ q
k
) t] − 1] ,
leading to
∂X
∂p
k
= P
k
· R
k
· q
k
·
exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
· [(p
k
+ q
k
) · t + exp [− (p
k
+ q
k
) t] − 1] ,
Now, we evaluate the derivative
∂X
′
∂p
k
:
∂X
′
∂p
k
= (P
k
· R
k
· q
k
)
′
·
∂
∂p
k
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
+
+ P
k
· R
k
· q
k
·
∂
∂p
k
(p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
−
− P
k
· R
k
· q
k
·
∂
∂p
k
q
′
k
+ [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t]) · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
=
= (P
k
· R
k
· q
k
)
′
·
exp [− (p
k
+ q
k
) t] · [(p
k
+ q
k
) · t + exp [− (p
k
+ q
k
) t] − 1]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
+
+ P
k
· R
k
· q
k
·
∂
∂p
k
(p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
−
− P
k
· R
k
· q
k
·
∂
∂p
k
q
′
k
+ [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t]) · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
.
Next, we evaluate
∂
∂p
k
(p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
=
=
exp [− (p
k
+ q
k
) t] + (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [− (p
k
+ q
k
) t] · (−t)
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
−
−
(p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
· exp [− (p
k
+ q
k
) t] · (1 − p
k
t) =
=
exp [− (p
k
+ q
k
) t] · (1 − (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · t)
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
−
−
(p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [−2 (p
k
+ q
k
) t] · (1 − p
k
t)
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
.
and
∂
∂p
k
q
′
k
+ [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)] · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
=
=
[− (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) − p
k
] · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
+
+
[p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t]) · exp [− (p
k
+ q
k
) t] · (−t)
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
−
−
q
′
k
+ [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)] · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
4
×
×
∂
∂p
k
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
=
=
exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
×
× · ([p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)] · t − (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) − p
k
) −
−
q
′
k
+ [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)] · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
4
×
× 2 (q
k
+ p
k
· exp [− (p
k
+ q
k
) t]) ·
∂
∂p
k
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t]) =
=
exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
×
× · ([p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)] · t − (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) − p
k
) −
−
q
′
k
+ [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)] · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
4
×
× 2 (q
k
+ p
k
· exp [− (p
k
+ q
k
) t]) (1 − p
k
· t) .
Thus, eventually,
∂X
′
∂p
k
= (P
k
· R
k
· q
k
)
′
·
exp [− (p
k
+ q
k
) t] · [(p
k
+ q
k
) · t + exp [− (p
k
+ q
k
) t] − 1]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
+
+ P
k
· R
k
· q
k
·
exp [− (p
k
+ q
k
) t] · (1 − (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · t)
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
−
− P
k
· R
k
· q
k
·
(p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [−2 (p
k
+ q
k
) t] · (1 − p
k
t)
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
−
− P
k
· R
k
· q
k
·
exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
×
× · ([p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)] · t − (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) − p
k
) +
+ P
k
· R
k
· q
k
·
q
′
k
+ [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)] · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
4
×
× 2 (q
k
+ p
k
· exp [− (p
k
+ q
k
) t]) (1 − p
k
· t) .
Similar steps as above will lead us to the following expressions:
∂X
∂q
k
=
∂
∂q
k
P
k
· R
k
· q
k
·
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
=
= P
k
· R
k
·
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
+ P
k
· R
k
· q
k
·
∂
∂p
k
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
,
in which
∂
∂q
k
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
=
1
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
∂
∂q
k
[1 − exp [− (p
k
+ q
k
) t]] +
+ (1 − exp [− (p
k
+ q
k
) t]) ·
∂
∂q
k
1
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
=
= −
exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
·
∂
∂q
k
[− (p
k
+ q
k
) t] −
−
1 − exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
·
∂
∂q
k
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t]) =
=
t · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
−
−
1 − exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
·
1 + p
k
·
∂ exp [− (p
k
+ q
k
) t]
∂q
k
=
=
t · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
−
−
(1 − exp [− (p
k
+ q
k
) t])
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
·
1 + p
k
· exp [− (p
k
+ q
k
) t] ·
∂
∂q
k
[− (p
k
+ q
k
) t]
=
=
t · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
−
(1 − exp [− (p
k
+ q
k
) t]) · (1 − p
k
· t · exp [− (p
k
+ q
k
) t])
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
,
leading to
∂X
∂q
k
= P
k
· R
k
·
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
+ P
k
· R
k
· q
k
·
t · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
−
− P
k
· R
k
· q
k
·
(1 − exp [− (p
k
+ q
k
) t]) · (1 − p
k
· t · exp [− (p
k
+ q
k
) t])
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
.
Now, we evaluate the derivative
∂X
′
∂q
k
:
∂X
′
∂q
k
= (P
k
· R
k
)
′
·
∂
∂q
k
q
k
·
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
+
+ P
k
· R
k
· q
′
k
·
∂
∂q
k
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
+
+ P
k
· R
k
·
∂
∂q
k
q
k
·
(p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
−
− P
k
· R
k
·
∂
∂q
k
q
k
·
q
′
k
+ [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)] · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
=
= (P
k
· R
k
)
′
·
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
+ (P
k
· R
k
)
′
· q
k
·
∂
∂q
k
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
+
+ P
k
· R
k
· q
′
k
·
∂
∂q
k
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
+
+ P
k
· R
k
·
(p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
+
+ P
k
· R
k
· q
k
·
∂
∂q
k
(p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
−
− P
k
· R
k
·
q
′
k
+ [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)] · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
−
− P
k
· R
k
· q
k
·
∂
∂q
k
q
′
k
+ [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)] · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
or
∂X
′
∂q
k
= (P
k
· R
k
)
′
·
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
+
+ (P
k
· R
k
· q
k
)
′
·
t · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
−
− (P
k
· R
k
· q
k
)
′
·
(1 − exp [− (p
k
+ q
k
) t]) · (1 − p
k
· t · exp [− (p
k
+ q
k
) t])
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
+
+ P
k
· R
k
·
(p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
+
+ P
k
· R
k
· q
k
·
∂
∂q
k
(p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
−
− P
k
· R
k
·
q
′
k
+ [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)] · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
−
− P
k
· R
k
· q
k
·
∂
∂q
k
q
′
k
+ [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)] · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
.
Furthermore,
∂
∂q
k
(p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
=
=
exp [− (p
k
+ q
k
) t] − (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [− (p
k
+ q
k
) t] · t
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
−
−
(p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
·
∂
∂q
k
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t]) =
=
exp [− (p
k
+ q
k
) t] − (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [− (p
k
+ q
k
) t] · t
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
−
−
(p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [− (p
k
+ q
k
) t] · (1 − p
k
· t · exp [− (p
k
+ q
k
) t])
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
,
and
∂
∂q
k
q
′
k
+ [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)] · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
=
=
−p
k
· exp [− (p
k
+ q
k
) t] + [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)] · exp [− (p
k
+ q
k
) t] · (−t)
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
−
−
q
′
k
+ [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)] · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
4
·
∂
∂q
k
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
=
= −
(p
k
+ t · [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)]) · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
−
−
q
′
k
+ [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)] · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
4
· (q
k
+ p
k
· exp [− (p
k
+ q
k
) t]) ×
× 2 · (1 − p
k
· t · exp [− (p
k
+ q
k
) t]) .
Thence,
∂X
′
∂q
k
= (P
k
· R
k
)
′
·
1 − exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
+
+ (P
k
· R
k
· q
k
)
′
·
t · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
−
− (P
k
· R
k
· q
k
)
′
·
(1 − exp [− (p
k
+ q
k
) t]) · (1 − p
k
· t · exp [− (p
k
+ q
k
) t])
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
+
+ P
k
· R
k
·
(p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
+
+ P
k
· R
k
· q
k
·
exp [− (p
k
+ q
k
) t] − (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [− (p
k
+ q
k
) t] · t
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
−
− P
k
· R
k
· q
k
·
(p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [− (p
k
+ q
k
) t] · (1 − p
k
· t · exp [− (p
k
+ q
k
) t])
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
−
− P
k
· R
k
·
q
′
k
+ [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)] · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
−
+ P
k
· R
k
· q
k
·
(p
k
+ t · [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)]) · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
−
+ P
k
· R
k
· q
k
·
q
′
k
+ [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)] · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
4
×
× 2 (q
k
+ p
k
· exp [− (p
k
+ q
k
) t]) · (1 − p
k
· t · exp [− (p
k
+ q
k
) t]) .
Apparently,
∂X
∂p
′
k
=
∂X
∂q
′
k
= 0,
and
∂X
′
∂p
′
k
=
∂
∂p
′
k
"
n
X
k=1
P
k
· R
k
· q
k
·
(p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
#
−
−
∂
∂p
′
k
"
n
X
k=1
P
k
· R
k
· q
k
·
q
′
k
+ [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)] · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
#
=
= P
k
· R
k
· q
k
·
t · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
− P
k
· R
k
· q
k
·
(1 − p
k
· t) · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
=
=
P
k
· R
k
· q
k
· exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
[t · (q
k
+ p
k
· exp [− (p
k
+ q
k
) t]) − 1 + p
k
· t] =
=
P
k
· R
k
· q
k
· exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
[(q
k
+ p
k
) · t + p
k
· t · exp [− (p
k
+ q
k
) t] − 1] .
Therefore,
∂Π
BTC
∂p
′
k
=
1
X
·
∂X
∂p
′
k
=
=
1
X
·
P
k
· R
k
· q
k
· exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
[(q
k
+ p
k
) · t + p
k
· t · exp [− (p
k
+ q
k
) t] − 1] .
Similarly,
∂X
′
∂q
′
k
=
∂
∂p
′
k
"
n
X
k=1
P
k
· R
k
· q
k
·
(p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t) · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
#
−
−
∂
∂p
′
k
"
n
X
k=1
P
k
· R
k
· q
k
·
q
′
k
+ [p
′
k
− p
k
· (p
k
+ q
k
+ (p
′
k
+ q
′
k
) · t)] · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
#
=
= P
k
· R
k
· q
k
·
t · exp [− (p
k
+ q
k
) t]
q
k
+ p
k
· exp [− (p
k
+ q
k
) t]
− P
k
· R
k
· q
k
·
(1 − p
k
· t) · exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
=
=
P
k
· R
k
· q
k
· exp [− (p
k
+ q
k
) t]
(q
k
+ p
k
· exp [− (p
k
+ q
k
) t])
2
[(q
k
+ p
k
) · t + p
k
· t · exp [− (p
k
+ q
k
) t] − 1] ,
i.e.,
∂X
′
∂p
′
k
=
∂X
′
∂q
′
k
.
Frechet Model for Alpha
For the absorption rate α
k
we consider the following model based on the well-known Frechet
distribution:
α
k
= p
k
· q
k
· t
−p
k
−1
,
where p
k
and q
k
are to be estimated.
Then, Π
BTC
and ln P
BTC
will obtain the following form:
Π
BTC
=
∂
∂t
"
ln
n
X
k=1
P
k
· R
k
· p
k
· q
k
· t
−p
k
−1
!
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d
#
,
ln P
BTC
= ln
n
X
k=1
P
k
· R
k
· p
k
· q
k
· t
−p
k
−1
!
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d.
Frechet Derivations
Here we calculate the sensitivity derivatives of
Π
BTC
=
∂
∂t
"
ln
n
X
k=1
P
k
· R
k
· p
k
· q
k
· t
−p
k
−1
!
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d
#
,
and
ln P
BTC
= ln
n
X
k=1
P
k
· R
k
· p
k
· q
k
· t
−p
k
−1
!
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d.
with respect to p
k
and q
k
.
Apparently,
∂Π
BTC
∂p
k
=
∂
2
∂t∂p
k
"
ln
n
X
k=1
P
k
· R
k
· p
k
· q
k
· t
−p
k
−1
!#
and
∂ ln P
BTC
∂p
k
=
∂
∂p
k
ln
n
X
k=1
P
k
· R
k
· p
k
· q
k
· t
−p
k
−1
!
,
since other terms do not depend on p
k
and q
k
.
Taking into account that
∂
∂t
"
ln
n
X
k=1
P
k
· R
k
· p
k
· q
k
· t
−p
k
−1
!#
=
X
′
X
,
where
X =
n
X
k=1
P
k
· R
k
· p
k
· q
k
· t
−p
k
−1
, X
′
=
∂X
∂t
,
we have
X
′
=
∂
∂t
n
X
k=1
P
k
· R
k
· p
k
· q
k
· t
−p
k
−1
!
= −
n
X
k=1
P
k
· R
k
· p
k
· (p
k
+ 1) · q
k
· t
−p
k
−2
.
Let us now calculate the sensitivity derivative
∂Π
BTC
∂p
k
=
∂
∂p
k
X
′
X
=
∂X
′
∂p
k
·
1
X
+ X
′
·
∂
∂p
k
1
X
=
∂X
′
∂p
k
·
1
X
−
X
′
X
2
·
∂X
∂p
k
.
Evidently,
∂X
′
∂p
k
= −P
k
· R
k
· q
k
·
∂
∂p
k
p
k
· (p
k
+ 1) · t
−p
k
−2
.
Applying the multiplication derivative rule, we get
∂
∂p
k
p
k
· (p
k
+ 1) · t
−p
k
−2
= (2p
k
+ 1) · t
−p
k
−2
− p
k
· (p
k
+ 1) · t
−p
k
−2
· ln t =
= t
−p
k
−2
· [2p
k
+ 1 − p
k
· (p
k
+ 1) · ln t] .
Thus,
∂X
′
∂p
k
= −P
k
· R
k
· q
k
· t
−p
k
−2
· [2p
k
+ 1 − p
k
· (p
k
+ 1) · ln t] .
On the other hand,
∂X
∂p
k
= P
k
· R
k
· q
k
·
∂
∂p
k
p
k
· t
−p
k
−1
=
= P
k
· R
k
· q
k
·
t
−p
k
−1
− p
k
· t
−p
k
−1
· ln t
=
= P
k
· R
k
· q
k
· t
−p
k
−1
· [1 − p
k
· ln t] .
Substituting the above expressions into the sensitivity derivative, we finally obtain
∂Π
BTC
∂p
k
= −
P
k
· R
k
· q
k
X
· t
−p
k
−2
· [2p
k
+ 1 − p
k
· (p
k
+ 1) · ln t] −
−
X
′
X
2
· P
k
· R
k
· q
k
· t
−p
k
−1
· [1 − p
k
· ln t] .
Let us now proceed with calculating the derivative of Π
BTC
with respect to q
k
. To that aim,
notice that X and X
′
are linear in q
k
leading to
∂X
∂q
k
= P
k
· R
k
· p
k
· t
−p
k
−1
·
∂
∂q
k
(q
k
) = P
k
· R
k
· p
k
· t
−p
k
−1
,
and
∂X
′
∂q
k
= −P
k
· R
k
· t
−p
k
−2
· [2p
k
+ 1 − p
k
· (p
k
+ 1) · ln t] .
Similarly, in the case of q
k
, we have
∂Π
BTC
∂q
k
= −
P
k
· R
k
X
· t
−p
k
−2
· [2p
k
+ 1 − p
k
· (p
k
+ 1) · ln t] −
−
X
′
X
2
· P
k
· R
k
· p
k
· t
−p
k
−1
.
Let us now proceed to the sensitivity derivatives calculations for ln P
BTC
. Apparently,
∂ ln P
BTC
∂p
k
=
∂ ln X
∂p
k
=
1
X
·
∂X
∂p
k
=
1
X
· P
k
· R
k
· q
k
· t
−p
k
−1
· [1 − p
k
· ln t] .
Similarly,
∂ ln P
BTC
∂q
k
=
∂ ln X
∂q
k
=
1
X
·
∂X
∂q
k
=
1
X
· P
k
· R
k
· p
k
· t
−p
k
−1
.
Weibull Model for Alpha
For the absorption rate α
k
we consider the following model based on the well-known Weibull
distribution:
α
k
=
p
k
q
k
·
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
,
where p
k
and q
k
are to be estimated.
Then, Π
BTC
and ln P
BTC
will obtain the following form:
Π
BTC
=
∂
∂t
"
ln
n
X
k=1
P
k
· R
k
·
p
k
q
k
·
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
!
+
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d
#
,
ln P
BTC
= ln
n
X
k=1
P
k
· R
k
·
p
k
q
k
·
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
!
+
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d.
Weibull Derivations
Here we calculate the sensitivity derivatives of
Π
BTC
=
∂
∂t
"
ln
n
X
k=1
P
k
· R
k
·
p
k
q
k
·
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
!
+
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d
#
and
ln P
BTC
= ln
n
X
k=1
P
k
· R
k
·
p
k
q
k
·
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
!
+
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d.
with respect to p
k
and q
k
.
Let us start with the sensitivity with respect to p
k
. Apparently, only the first term in both
expressions above depends on p
k
. Therefore, we may dismiss other terms when calculating the
partial derivatives. In other words,
∂Π
BTC
∂p
k
=
∂
2
∂t∂p
k
"
ln
n
X
k=1
P
k
· R
k
·
p
k
q
k
·
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
!#
and
∂ ln P
BTC
∂p
k
=
∂
∂p
k
ln
n
X
k=1
P
k
· R
k
·
p
k
q
k
·
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
!
.
Taking into account that
∂
∂t
"
ln
n
X
k=1
P
k
· R
k
·
p
k
q
k
·
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
!#
=
X
′
X
,
where
X =
n
X
k=1
P
k
· R
k
·
p
k
q
k
·
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
and
X
′
=
∂X
∂t
,
we have
X
′
=
∂
∂t
n
X
k=1
P
k
· R
k
·
p
k
q
k
·
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
!
=
=
n
X
k=1
P
k
· R
k
·
p
k
q
k
·
∂
∂t
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
!
.
On the other hand,
∂
∂t
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
!
=
∂
∂t
"
t
q
k
p
k
−1
#
· exp
−
t
q
k
p
k
+
+
t
q
k
p
k
−1
·
∂
∂t
exp
−
t
q
k
p
k
=
p
k
− 1
q
k
·
t
q
k
p
k
−2
· exp
−
t
q
k
p
k
+
+
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
·
−
p
k
q
k
·
t
q
k
p
k
−1
!
=
=
p
k
− 1
q
k
·
t
q
k
p
k
−2
· exp
−
t
q
k
p
k
−
p
k
q
k
·
t
q
k
2p
k
−2
· exp
−
t
q
k
p
k
=
=
t
q
k
p
k
−2
· exp
−
t
q
k
p
k
·
p
k
− 1
q
k
−
p
k
q
k
·
t
q
k
p
k
.
Thus, finally,
X
′
=
n
X
k=1
P
k
· R
k
·
p
k
q
k
·
t
q
k
p
k
−2
· exp
−
t
q
k
p
k
·
p
k
− 1
q
k
−
p
k
q
k
·
t
q
k
p
k
.
Coming back to the sensitivity problem, let us now calculate the sensitivity derivative
∂Π
BTC
∂p
k
=
∂
∂p
k
X
′
X
=
∂X
′
∂p
k
·
1
X
+ X
′
·
∂
∂p
k
1
X
=
∂X
′
∂p
k
·
1
X
−
X
′
X
2
·
∂X
∂p
k
.
Apparently,
∂X
′
∂p
k
= P
k
· R
k
·
∂
∂p
k
p
k
q
k
·
t
q
k
p
k
−2
· exp
−
t
q
k
p
k
·
p
k
− 1
q
k
−
p
k
q
k
·
t
q
k
p
k
!
.
Applying the multiplication derivative rule, we get
∂
∂p
k
p
k
q
k
·
t
q
k
p
k
−2
· exp
−
t
q
k
p
k
·
p
k
− 1
q
k
−
p
k
q
k
·
t
q
k
p
k
!
=
=
∂
∂p
k
p
k
q
k
·
t
q
k
p
k
−2
· exp
−
t
q
k
p
k
·
p
k
− 1
q
k
−
p
k
q
k
·
t
q
k
p
k
+
+
p
k
q
k
·
∂
∂p
k
t
q
k
p
k
−2
!
· exp
−
t
q
k
p
k
·
p
k
− 1
q
k
−
p
k
q
k
·
t
q
k
p
k
+
+
p
k
q
k
·
t
q
k
p
k
−2
·
∂
∂p
k
exp
−
t
q
k
p
k
·
p
k
− 1
q
k
−
p
k
q
k
·
t
q
k
p
k
+
+
p
k
q
k
·
t
q
k
p
k
−2
· exp
−
t
q
k
p
k
·
∂
∂p
k
p
k
− 1
q
k
−
p
k
q
k
·
t
q
k
p
k
.
Since
∂
∂p
k
p
k
q
k
=
1
q
k
,
∂
∂p
k
t
q
k
p
k
−2
!
=
t
q
k
p
k
−2
· ln
t
q
k
,
∂
∂p
k
exp
−
t
q
k
p
k
= exp
−
t
q
k
p
k
·
−
t
q
k
p
k
· ln
t
q
k
=
= −
t
q
k
p
k
· ln
t
q
k
· exp
−
t
q
k
p
k
,
∂
∂p
k
p
k
− 1
q
k
−
p
k
q
k
·
t
q
k
p
k
=
1
q
k
−
1
q
k
·
t
q
k
p
k
−
p
k
q
k
·
t
q
k
p
k
· ln
t
q
k
=
=
1
q
k
−
1
q
k
·
t
q
k
p
k
·
1 − p
k
· ln
t
q
k
=
=
1
q
k
−
1
q
k
·
t
q
k
p
k
·
1 − ln
t
q
k
p
k
.
Combining all separate derivatives, we will obtain
∂
∂p
k
p
k
q
k
·
t
q
k
p
k
−2
· exp
−
t
q
k
p
k
·
p
k
− 1
q
k
−
p
k
q
k
·
t
q
k
p
k
!
=
=
1
q
k
·
t
q
k
p
k
−2
· exp
−
t
q
k
p
k
·
p
k
− 1
q
k
−
p
k
q
k
·
t
q
k
p
k
+
+
p
k
q
k
·
t
q
k
p
k
−2
· ln
t
q
k
· exp
−
t
q
k
p
k
·
p
k
− 1
q
k
−
p
k
q
k
·
t
q
k
p
k
−
−
p
k
q
k
·
t
q
k
p
k
−2
·
t
q
k
p
k
· ln
t
q
k
· exp
−
t
q
k
p
k
·
p
k
− 1
q
k
−
p
k
q
k
·
t
q
k
p
k
+
+
p
k
q
k
·
t
q
k
p
k
−2
· exp
−
t
q
k
p
k
·
1
q
k
−
1
q
k
·
t
q
k
p
k
·
1 − ln
t
q
k
p
k
=
=
t
q
k
p
k
−2
· exp
−
t
q
k
p
k
·
"
1
q
k
·
p
k
− 1
q
k
−
p
k
q
k
·
t
q
k
p
k
+
+
1
q
k
· ln
t
q
k
p
k
·
p
k
− 1
q
k
−
p
k
q
k
·
t
q
k
p
k
−
−
1
q
k
·
t
q
k
p
k
· ln
t
q
k
p
k
·
p
k
− 1
q
k
p
k
q
k
·
t
q
k
p
k
+
+
p
k
q
k
·
1
q
k
·
1 −
t
q
k
p
k
·
1 − ln
t
q
k
p
k
#
=
=
1
q
k
·
t
q
k
p
k
−2
· exp
−
t
q
k
p
k
·
"
p
k
q
k
·
1 −
t
q
k
p
k
·
1 − ln
t
q
k
p
k
+
+
p
k
− 1
q
k
−
p
k
q
k
·
t
q
k
p
k
·
1 + ln
t
q
k
p
k
−
t
q
k
p
k
· ln
t
q
k
p
k
#
.
Thus,
∂X
′
∂p
k
=
P
k
· R
k
q
k
·
t
q
k
p
k
−2
· exp
−
t
q
k
p
k
·
"
p
k
q
k
·
1 −
t
q
k
p
k
·
1 − ln
t
q
k
p
k
+
+
p
k
− 1
q
k
−
p
k
q
k
·
t
q
k
p
k
·
1 + ln
t
q
k
p
k
−
t
q
k
p
k
· ln
t
q
k
p
k
#
.
Now, compute the following derivative:
∂X
∂p
k
= P
k
· R
k
·
∂
∂p
k
p
k
q
k
·
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
!
=
= P
k
· R
k
·
"
1
q
k
·
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
+
+
p
k
q
k
·
t
q
k
p
k
−1
· ln
t
q
k
· exp
−
t
q
k
p
k
+
+
p
k
q
k
·
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
·
−
t
q
k
p
k
· ln
t
q
k
#
=
=
P
k
· R
k
q
k
·
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
·
"
1 + ln
t
q
k
p
k
·
1 −
t
q
k
p
k
#
.
Substituting the above expressions into the sensitivity derivative, we finally obtain
∂Π
BTC
∂p
k
=
1
X
·
P
k
· R
k
q
k
·
t
q
k
p
k
−2
· exp
−
t
q
k
p
k
·
"
p
k
q
k
·
1 −
t
q
k
p
k
·
1 − ln
t
q
k
p
k
+
+
p
k
− 1
q
k
−
p
k
q
k
·
t
q
k
p
k
·
1 + ln
t
q
k
p
k
−
t
q
k
p
k
· ln
t
q
k
p
k
#
−
−
X
′
X
2
·
P
k
· R
k
q
k
·
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
·
"
1 + ln
t
q
k
p
k
·
1 −
t
q
k
p
k
#
.
Before proceeding to computing of derivatives with respect to q
k
, let us simplify the expres-
sions for X and X
′
. Namely,
X =
n
X
k=1
P
k
· R
k
·
p
k
q
k
·
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
=
=
n
X
k=1
P
k
· R
k
· p
k
· q
−1
k
· t
p
k
−1
· q
−p
k
+1
k
· exp
−t
p
k
· q
−p
k
k
=
=
n
X
k=1
P
k
· R
k
· p
k
· t
p
k
−1
· q
−p
k
k
· exp
−t
p
k
· q
−p
k
k
,
X
′
=
n
X
k=1
P
k
· R
k
·
p
k
q
k
·
t
q
k
p
k
−2
· exp
−
t
q
k
p
k
·
p
k
− 1
q
k
−
p
k
q
k
·
t
q
k
p
k
=
=
n
X
k=1
P
k
· R
k
· p
k
· q
−1
k
· t
p
k
−2
· q
−p
k
+2
k
· exp
−t
p
k
· q
−p
k
k
· q
−1
k
·
p
k
− 1 − p
k
· t
p
k
· q
−p
k
k
=
=
n
X
k=1
P
k
· R
k
· p
k
· t
p
k
−2
· q
−p
k
k
· exp
−t
p
k
· q
−p
k
k
·
p
k
− 1 − p
k
· t
p
k
· q
−p
k
k
.
Thus,
∂X
∂q
k
= P
k
· R
k
· p
k
· t
p
k
−1
·
∂
∂q
k
q
−p
k
k
· exp
−t
p
k
· q
−p
k
k
=
= P
k
· R
k
· p
k
· t
p
k
−1
·
∂q
−p
k
k
∂q
k
· exp
−t
p
k
· q
−p
k
k
+ q
−p
k
k
·
∂
∂q
k
exp
−t
p
k
· q
−p
k
k
=
= P
k
· R
k
· p
k
· t
p
k
−1
· exp
−t
p
k
· q
−p
k
k
·
"
− p
k
q
−p
k
−1
k
+ q
−p
k
k
·
−p
k
· t
p
k
· q
−p
k
−1
k
#
=
= −P
k
· R
k
· p
2
k
· t
p
k
−1
· q
−p
k
−1
k
· exp
−t
p
k
· q
−p
k
k
·
"
1 + q
−p
k
k
· t
p
k
#
=
= −P
k
· R
k
· p
2
k
·
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
·
"
1 +
t
q
k
p
k
#
,
and
∂X
′
∂q
k
= P
k
· R
k
· p
k
· t
p
k
−2
·
∂
∂q
k
q
−p
k
k
· exp
−t
p
k
· q
−p
k
k
·
p
k
− 1 − p
k
· t
p
k
· q
−p
k
k
=
= −P
k
· R
k
· p
2
k
·
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
·
"
1 +
t
q
k
p
k
#
·
p
k
·
1 −
t
q
k
p
k
− 1
+
+ P
k
· R
k
· p
3
k
· q
k
·
t
q
k
2p
k
−2
· exp
−
t
q
k
p
k
.
Similarly, in the case of q
k
, we have
∂Π
BTC
∂q
k
=
∂
∂q
k
X
′
X
=
∂X
′
∂q
k
·
1
X
−
X
′
X
2
·
∂X
∂q
k
=
= −
P
k
· R
k
· p
2
k
X
·
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
·
"
1 +
t
q
k
p
k
#
·
p
k
·
1 −
t
q
k
p
k
− 1
+
+
P
k
· R
k
· p
3
k
· q
k
X
·
t
q
k
2p
k
−2
· exp
−
t
q
k
p
k
+
+
X
′
· P
k
· R
k
· p
2
k
X
2
·
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
·
"
1 +
t
q
k
p
k
#
.
Let us now proceed to the sensitivity derivatives calculations for ln P
BTC
. Apparently,
∂ ln P
BTC
∂p
k
=
∂ ln X
∂p
k
=
1
X
·
∂X
∂p
k
=
=
P
k
· R
k
X · q
k
·
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
·
"
1 + ln
t
q
k
p
k
·
1 −
t
q
k
p
k
#
.
Similarly,
∂ ln P
BTC
∂q
k
=
∂ ln X
∂q
k
=
1
X
·
∂X
∂q
k
=
= −
P
k
· R
k
· p
2
k
X
·
t
q
k
p
k
−1
· exp
−
t
q
k
p
k
·
"
1 +
t
q
k
p
k
#
.
Gumbel Model for Alpha
For the absorption rate α
k
we consider the following model based on the well-known Gumbel
distribution:
α
k
= p
k
· q
k
· t
−p
k
−1
· exp
−q
k
t
−p
k
,
where p
k
and q
k
are to be estimated.
Then, Π
BTC
and ln P
BTC
will obtain the following form:
Π
BTC
=
∂
∂t
"
ln
n
X
k=1
P
k
· R
k
· p
k
· q
k
· t
−p
k
−1
· exp
−q
k
t
−p
k
!
+
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d
#
,
ln P
BTC
= ln
n
X
k=1
P
k
· R
k
· p
k
· q
k
· t
−p
k
−1
· exp
−q
k
t
−p
k
!
+
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d.
Gumbel Derivations
Here we calculate the sensitivity derivatives of
Π
BTC
=
∂
∂t
"
ln
n
X
k=1
P
k
· R
k
· p
k
· q
k
· t
−p
k
−1
· exp
−q
k
t
−p
k
!
+
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d
#
,
and
ln P
BTC
= ln
n
X
k=1
P
k
· R
k
· p
k
· q
k
· t
−p
k
−1
· exp
−q
k
t
−p
k
!
+
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d.
with respect to p
k
and q
k
.
Apparently,
∂Π
BTC
∂p
k
=
∂
2
∂t∂p
k
"
ln
n
X
k=1
P
k
· R
k
· p
k
· q
k
· t
−p
k
−1
· exp
−q
k
t
−p
k
!#
and
∂ ln P
BTC
∂p
k
=
∂
∂p
k
ln
n
X
k=1
P
k
· R
k
· p
k
· q
k
· t
−p
k
−1
· exp
−q
k
t
−p
k
!
,
since other terms do not depend on p
k
and q
k
.
Taking into account that
∂
∂t
"
ln
n
X
k=1
P
k
· R
k
· p
k
· q
k
· t
−p
k
−1
· exp
−q
k
t
−p
k
!#
=
X
′
X
,
where
X =
n
X
k=1
P
k
· R
k
· p
k
· q
k
· t
−p
k
−1
· exp
−q
k
t
−p
k
and
X
′
=
∂X
∂t
,
we have
X
′
=
∂
∂t
n
X
k=1
P
k
· R
k
· p
k
· q
k
· t
−p
k
−1
· exp
−q
k
t
−p
k
!
=
=
n
X
k=1
P
k
· R
k
· p
k
· q
k
·
∂t
−p
k
−1
∂t
· exp
−q
k
t
−p
k
+ t
−p
k
−1
·
∂
∂t
exp
−q
k
t
−p
k
=
=
n
X
k=1
P
k
· R
k
· p
k
· q
k
·
(−p
k
− 1) · t
−p
k
−2
+ t
−p
k
−1
·
p
k
· q
k
· t
−p
k
−1
· exp
−q
k
t
−p
k
=
=
n
X
k=1
P
k
· R
k
· p
k
· q
k
·
−p
k
− 1 + p
k
· q
k
· t
−p
k
· t
−p
k
−2
· exp
−q
k
t
−p
k
.
Thus, finally,
X
′
=
n
X
k=1
P
k
· R
k
· p
k
· q
k
·
−p
k
− 1 + p
k
· q
k
· t
−p
k
· t
−p
k
−2
· exp
−q
k
t
−p
k
.
Coming back to the sensitivity problem, let us now calculate the sensitivity derivative
∂Π
BTC
∂p
k
=
∂
∂p
k
X
′
X
=
∂X
′
∂p
k
·
1
X
+ X
′
·
∂
∂p
k
1
X
=
∂X
′
∂p
k
·
1
X
−
X
′
X
2
·
∂X
∂p
k
.
Evidently,
∂X
′
∂p
k
= P
k
· R
k
· q
k
·
∂
∂p
k
p
k
·
−p
k
− 1 + p
k
· q
k
· t
−p
k
· t
−p
k
−2
· exp
−q
k
t
−p
k
.
Applying the multiplication derivative rule, we get
∂
∂p
k
p
k
·
−p
k
− 1 + p
k
· q
k
· t
−p
k
· t
−p
k
−2
· exp
−q
k
t
−p
k
=
=
−p
k
− 1 + p
k
· q
k
· t
−p
k
· t
−p
k
−2
· exp
−q
k
t
−p
k
+
+ p
k
·
∂
∂p
k
−p
k
− 1 + p
k
· q
k
· t
−p
k
· t
−p
k
−2
· exp
−q
k
t
−p
k
+
+ p
k
·
−p
k
− 1 + p
k
· q
k
· t
−p
k
·
∂t
−p
k
−2
∂p
k
· exp
−q
k
t
−p
k
+
+ p
k
·
−p
k
− 1 + p
k
· q
k
· t
−p
k
· t
−p
k
−2
·
∂ exp [−q
k
t
−p
k
]
∂p
k
=
=
−p
k
− 1 + p
k
· q
k
· t
−p
k
· t
−p
k
−2
· exp
−q
k
t
−p
k
+
+ p
k
·
−1 + q
k
· t
−p
k
− p
k
· q
k
· t
−p
k
· ln t
· t
−p
k
−2
· exp
−q
k
t
−p
k
−
− p
k
·
−p
k
− 1 + p
k
· q
k
· t
−p
k
· t
−p
k
−2
· ln t · exp
−q
k
t
−p
k
+
+ p
k
·
−p
k
− 1 + p
k
· q
k
· t
−p
k
· t
−p
k
−2
· exp
−q
k
t
−p
k
· q
k
· t
−p
k
· ln t
or
∂
∂p
k
p
k
·
−p
k
− 1 + p
k
· q
k
· t
−p
k
· t
−p
k
−2
· exp
−q
k
t
−p
k
=
= t
−p
k
−2
· exp
−q
k
t
−p
k
·
"
−p
k
− 1 + p
k
· q
k
· t
−p
k
+
+ p
k
·
−1 + q
k
· t
−p
k
− p
k
· q
k
· t
−p
k
· ln t
−
− p
k
·
−p
k
− 1 + p
k
· q
k
· t
−p
k
+
+ p
k
·
−p
k
− 1 + p
k
· q
k
· t
−p
k
· q
k
· t
−p
k
· ln t
#
=
= t
−p
k
−2
· exp
−q
k
t
−p
k
·
"
p
k
·
−1 + q
k
· t
−p
k
− p
k
· q
k
· t
−p
k
· ln t
+
+
−p
k
− 1 + p
k
· q
k
· t
−p
k
·
1 − p
k
+ p
k
· q
k
· t
−p
k
· ln t
#
.
Thus, combining all separate derivatives, we will obtain
∂X
′
∂p
k
= P
k
· R
k
· q
k
· t
−p
k
−2
· exp
−q
k
t
−p
k
·
"
p
k
·
−1 + q
k
· t
−p
k
− p
k
· q
k
· t
−p
k
· ln t
+
+
−p
k
− 1 + p
k
· q
k
· t
−p
k
·
1 − p
k
+ p
k
· q
k
· t
−p
k
· ln t
#
.
Now, compute the following derivative:
∂X
∂p
k
= P
k
· R
k
· q
k
·
∂
∂p
k
p
k
· t
−p
k
−1
· exp
−q
k
t
−p
k
=
= P
k
· R
k
· q
k
·
t
−p
k
−1
· exp
−q
k
t
−p
k
− p
k
· t
−p
k
−1
· ln t · exp
−q
k
t
−p
k
+
+ p
k
· t
−p
k
−1
· exp
−q
k
t
−p
k
· q
k
· t
−p
k
· ln t
=
= P
k
· R
k
· q
k
· t
−p
k
−1
· exp
−q
k
t
−p
k
·
1 − p
k
· ln t + p
k
· q
k
· t
−p
k
· ln t
=
= P
k
· R
k
· q
k
· t
−p
k
−1
· exp
−q
k
t
−p
k
·
1 − p
k
· ln t
1 − q
k
· t
−p
k
.
Substituting the above expressions into the sensitivity derivative, we finally obtain
∂Π
BTC
∂p
k
=
P
k
· R
k
· q
k
X
· t
−p
k
−2
· exp
−q
k
t
−p
k
·
"
p
k
·
−1 + q
k
· t
−p
k
− p
k
· q
k
· t
−p
k
· ln t
+
+
−p
k
− 1 + p
k
· q
k
· t
−p
k
·
1 − p
k
+ p
k
· q
k
· t
−p
k
· ln t
#
−
−
X
′
X
2
· P
k
· R
k
· q
k
· t
−p
k
−1
· exp
−q
k
t
−p
k
·
1 − p
k
· ln t
1 − q
k
· t
−p
k
.
Let us now proceed with calculating the derivative of Π
BTC
with respect to q
k
. To that aim,
let us first compute the analogous derivatives
∂X
∂q
k
= P
k
· R
k
· p
k
· t
−p
k
−1
·
∂
∂q
k
q
k
· exp
−q
k
t
−p
k
=
= P
k
· R
k
· p
k
· t
−p
k
−1
·
exp
−q
k
t
−p
k
+ exp
−q
k
t
−p
k
·
−t
−p
k
=
= P
k
· R
k
· p
k
· t
−p
k
−1
· exp
−q
k
t
−p
k
·
1 − t
−p
k
,
and
∂X
′
∂q
k
= P
k
· R
k
· p
k
· t
−p
k
−2
·
∂
∂q
k
q
k
·
−p
k
− 1 + p
k
· q
k
· t
−p
k
· exp
−q
k
t
−p
k
=
= P
k
· R
k
· p
k
· t
−p
k
−2
·
−p
k
− 1 + p
k
· q
k
· t
−p
k
· exp
−q
k
t
−p
k
+
+ q
k
· p
k
· t
−p
k
· exp
−q
k
t
−p
k
− q
k
·
−p
k
− 1 + p
k
· q
k
· t
−p
k
· exp
−q
k
t
−p
k
· t
−p
k
=
= P
k
· R
k
· p
k
· t
−p
k
−2
· exp
−q
k
t
−p
k
−p
k
− 1 + p
k
· q
k
· t
−p
k
·
1 − t
−p
k
+
+ q
k
· p
k
· t
−p
k
.
Similarly, in the case of q
k
, we have
∂Π
BTC
∂q
k
=
∂
∂q
k
X
′
X
=
∂X
′
∂q
k
·
1
X
−
X
′
X
2
·
∂X
∂q
k
=
=
P
k
· R
k
· p
k
X
· t
−p
k
−2
· exp
−q
k
t
−p
k
−p
k
− 1 + p
k
· q
k
· t
−p
k
·
1 − t
−p
k
+
+ q
k
· p
k
· t
−p
k
−
X
′
X
2
· P
k
· R
k
· p
k
· t
−p
k
−1
· exp
−q
k
t
−p
k
·
1 − t
−p
k
.
Let us now proceed to the sensitivity derivatives calculations for ln P
BTC
. Apparently,
∂ ln P
BTC
∂p
k
=
∂ ln X
∂p
k
=
1
X
·
∂X
∂p
k
=
=
1
X
· P
k
· R
k
· q
k
· t
−p
k
−1
· exp
−q
k
t
−p
k
·
1 − p
k
· ln t
1 − q
k
· t
−p
k
.
Similarly,
∂ ln P
BTC
∂q
k
=
∂ ln X
∂q
k
=
1
X
·
∂X
∂q
k
=
=
1
X
· P
k
· R
k
· p
k
· t
−p
k
−1
· exp
−q
k
t
−p
k
·
1 − t
−p
k
.
Shifted Gompertz Model for Alpha
Consider now the following particular model for the absorption rate α
k
corresponding to the
well-known shifted Gompertz distribution:
α
k
= p
k
· exp [−p
k
t − q
k
exp (−p
k
t)] · (1 + q
k
· [1 − exp (−p
k
t)]) ,
where p
k
and q
k
are to be estimated.
Then, Π
BTC
and ln P
BTC
will obtain the following form:
Π
BTC
=
∂
∂t
"
ln
n
X
k=1
P
k
· p
k
· exp [−p
k
t − q
k
exp (−p
k
t)] · (1 + q
k
· [1 − exp (−p
k
t)]) · R
k
!
+
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d
#
,
ln P
BTC
= ln
n
X
k=1
P
k
· p
k
· exp [−p
k
t − q
k
exp (−p
k
t)] · (1 + q
k
· [1 − exp (−p
k
t)]) · R
k
!
+
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d.
Shifted Gompertz Derivations
In this appendix, we will consider the sensitivity of
Π
BTC
=
∂
∂t
"
ln
n
X
k=1
P
k
· p
k
· exp [−p
k
t − q
k
exp (−p
k
t)] · (1 + q
k
· [1 − exp (−p
k
t)]) · R
k
!
+
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d
#
and
ln P
BTC
= ln
n
X
k=1
P
k
· p
k
· exp [−p
k
t − q
k
exp (−p
k
t)] · (1 + q
k
· [1 − exp (−p
k
t)]) · R
k
!
+
+ ln
1
m
m
X
j=1
T
′
j
!
− ln b − ln h + ln d
with respect to p
k
and q
k
.
Let us start with the sensitivity with respect to p
k
. Apparently, only the first term in both
expressions above depends on p
k
. Therefore, we may dismiss other terms when calculating the
partial derivatives. In other words,
∂Π
BTC
∂p
k
=
∂
2
∂t∂p
k
"
ln
n
X
k=1
P
k
· p
k
· exp [−p
k
t − q
k
exp (−p
k
t)] · (1 + q
k
· [1 − exp (−p
k
t)]) · R
k
!#
and
∂ ln P
BTC
∂p
k
=
∂
∂p
k
ln
n
X
k=1
P
k
· p
k
· exp [−p
k
t − q
k
exp (−p
k
t)] · (1 + q
k
· [1 − exp (−p
k
t)]) · R
k
!
Taking into account that
∂
∂t
"
ln
n
X
k=1
P
k
· p
k
· exp [−p
k
t − q
k
exp (−p
k
t)] · (1 + q
k
· [1 − exp (−p
k
t)]) · R
k
!#
=
X
′
X
,
where
X =
n
X
k=1
P
k
· p
k
· exp [−p
k
t − q
k
exp (−p
k
t)] · (1 + q
k
· [1 − exp (−p
k
t)]) · R
k
and
X
′
=
∂X
∂t
,
we have
X
′
=
∂
∂t
"
n
X
k=1
P
k
· p
k
· exp [−p
k
t − q
k
exp (−p
k
t)] · (1 + q
k
· [1 − exp (−p
k
t)]) · R
k
#
=
=
n
X
k=1
P
k
· p
k
· R
k
·
∂
∂t
h
exp [−p
k
t − q
k
exp (−p
k
t)] · (1 + q
k
· [1 − exp (−p
k
t)])
i
.
On the other hand,
∂
∂t
h
exp [−p
k
t − q
k
exp (−p
k
t)] · (1 + q
k
· [1 − exp (−p
k
t)])
i
=
=
∂
∂t
h
exp [−p
k
t − q
k
exp (−p
k
t)]
i
· (1 + q
k
· [1 − exp (−p
k
t)]) +
+ exp [−p
k
t − q
k
exp (−p
k
t)] ·
∂
∂t
h
(1 + q
k
· [1 − exp (−p
k
t)])
i
=
= exp [−p
k
t − q
k
exp (−p
k
t)] ·
∂
∂t
− p
k
t − q
k
exp (−p
k
t)
· (1 + q
k
· [1 − exp (−p
k
t)]) +
+ exp [−p
k
t − q
k
exp (−p
k
t)] ·
−q
k
·
∂
∂t
h
exp (−p
k
t)
i
=
= exp [−p
k
t − q
k
exp (−p
k
t)] ·
− p
k
− q
k
exp (−p
k
t) · (−p
k
)
· (1 + q
k
· [1 − exp (−p
k
t)]) +
+ exp [−p
k
t − q
k
exp (−p
k
t)] · (p
k
· q
k
· exp (−p
k
t)) =
= −p
k
· exp [−p
k
t − q
k
exp (−p
k
t)] ·
1 − q
k
exp (−p
k
t)
· (1 + q
k
· [1 − exp (−p
k
t)]) +
+ exp [−p
k
t − q
k
exp (−p
k
t)] · p
k
· q
k
· exp (−p
k
t) =
= p
k
· exp [−p
k
t − q
k
exp (−p
k
t)] ·
"
q
k
· exp (−p
k
t) −
−
1 − q
k
exp (−p
k
t)
· (1 + q
k
· [1 − exp (−p
k
t)])
#
.
Simplifying the expression in the last bracket, we derive
q
k
· exp (−p
k
t) −
1 − q
k
exp (−p
k
t)
· (1 + q
k
· [1 − exp (−p
k
t)]) =
= q
k
· exp (−p
k
t) − 1 − q
k
· [1 − exp (−p
k
t)] + q
k
· exp (−p
k
t) +
+ q
2
k
· exp (−p
k
t) · [1 − exp (−p
k
t)] =
= 2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1.
Therefore, finally,
X
′
= −
n
X
k=1
P
k
· R
k
· p
2
k
· exp [−p
k
t − q
k
exp (−p
k
t)] ×
×
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
.
Coming back to the sensitivity problem, let us now calculate the sensitivity derivative
∂Π
BTC
∂p
k
=
∂
∂p
k
X
′
X
=
∂X
′
∂p
k
·
1
X
+ X
′
·
∂
∂p
k
1
X
=
∂X
′
∂p
k
·
1
X
−
X
′
X
2
·
∂X
∂p
k
.
Apparently,
∂X
′
∂p
k
= −P
k
· R
k
·
∂
∂p
k
p
2
k
· exp [−p
k
t − q
k
exp (−p
k
t)] ×
×
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
.
Applying the multiplication derivative rule, we get
∂
∂p
k
p
2
k
· exp [−p
k
t − q
k
exp (−p
k
t)] ×
×
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
=
=
∂p
2
k
∂p
k
· exp [−p
k
t − q
k
exp (−p
k
t)] ×
×
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
+
+ p
2
k
·
∂ exp [−p
k
t − q
k
exp (−p
k
t)]
∂p
k
×
×
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
+
+ p
2
k
· exp [−p
k
t − q
k
exp (−p
k
t)] ×
×
∂
∂p
k
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
=
= 2p
k
· exp [−p
k
t − q
k
exp (−p
k
t)] ×
×
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
+
+ p
2
k
· exp [−p
k
t − q
k
exp (−p
k
t)] · [−t − q
k
exp (−p
k
t) · (−t)] ×
×
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
+
+ p
2
k
· exp [−p
k
t − q
k
exp (−p
k
t)] ×
×
2q
k
· exp (−p
k
t) · (−t) − q
k
· [− exp (−p
k
t) · (−t)] · (1 − q
k
· exp (−p
k
t)) −
− q
k
· [1 − exp (−p
k
t)] · (−q
k
· exp (−p
k
t) · (−t))
=
= 2p
k
· exp [−p
k
t − q
k
exp (−p
k
t)] ×
×
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
−
− p
2
k
t · exp [−p
k
t − q
k
exp (−p
k
t)] · [1 − q
k
exp (−p
k
t)] ×
×
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
+
+ p
2
k
· exp [−p
k
t − q
k
exp (−p
k
t)] ×
×
− 2q
k
t · exp (−p
k
t) − q
k
t · exp (−p
k
t) · (1 − q
k
· exp (−p
k
t)) −
− q
2
k
t · exp (−p
k
t) · [1 − exp (−p
k
t)]
=
= 2p
k
· exp [−p
k
t − q
k
exp (−p
k
t)] ×
×
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
−
− p
2
k
t · exp [−p
k
t − q
k
exp (−p
k
t)] · [1 − q
k
exp (−p
k
t)] ×
×
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
+
+ p
2
k
· exp [−p
k
t − q
k
exp (−p
k
t)] ×
× q
k
t · exp (−p
k
t) ·
− 2 − (1 − q
k
· exp (−p
k
t)) − q
k
· [1 − exp (−p
k
t)]
=
= 2p
k
· exp [−p
k
t − q
k
exp (−p
k
t)] ×
×
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
−
− p
2
k
t · exp [−p
k
t − q
k
exp (−p
k
t)] · [1 − q
k
exp (−p
k
t)] ×
×
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
−
− p
2
k
· q
k
t · exp (−p
k
t) · exp [−p
k
t − q
k
exp (−p
k
t)] ×
×
2 + (1 − q
k
· exp (−p
k
t)) + q
k
· [1 − exp (−p
k
t)]
.
Thus,
∂X
′
∂p
k
= −P
k
· R
k
· 2p
k
· exp [−p
k
t − q
k
exp (−p
k
t)] ×
×
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
−
− p
2
k
t · exp [−p
k
t − q
k
exp (−p
k
t)] · [1 − q
k
exp (−p
k
t)] ×
×
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
−
− p
2
k
· q
k
t · exp (−p
k
t) · exp [−p
k
t − q
k
exp (−p
k
t)] ×
×
2 + (1 − q
k
· exp (−p
k
t)) + q
k
· [1 − exp (−p
k
t)]
.
Now, compute the following derivative:
∂X
∂p
k
= P
k
· R
k
·
∂
∂p
k
[p
k
· exp [−p
k
t − q
k
exp (−p
k
t)] · (1 + q
k
· [1 − exp (−p
k
t)])] =
= P
k
· R
k
·
h
exp [−p
k
t − q
k
exp (−p
k
t)] · (1 + q
k
· [1 − exp (−p
k
t)]) +
+ p
k
· exp [−p
k
t − q
k
exp (−p
k
t)] · [−t − q
k
exp (−p
k
t) · (−t)] ×
× (1 + q
k
· [1 − exp (−p
k
t)]) + p
k
· exp [−p
k
t − q
k
exp (−p
k
t)] · q
k
· [− exp (−p
k
t) · (−t)]
i
=
= P
k
· R
k
· exp [−p
k
t − q
k
exp (−p
k
t)] ·
h
1 + q
k
· [1 − exp (−p
k
t)] −
− p
k
t · [1 − q
k
exp (−p
k
t)] · (1 + q
k
· [1 − exp (−p
k
t)]) + p
k
· q
k
t · exp (−p
k
t)
i
.
Substituting the above expressions into the sensitivity derivative, we finally obtain
∂Π
BTC
∂p
k
= −
P
k
· R
k
· 2p
k
· exp [−p
k
t − q
k
exp (−p
k
t)]
X
×
×
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
−
− p
2
k
t · exp [−p
k
t − q
k
exp (−p
k
t)] · [1 − q
k
exp (−p
k
t)] ×
×
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
−
− p
2
k
· q
k
t · exp (−p
k
t) · exp [−p
k
t − q
k
exp (−p
k
t)] ×
×
2 + (1 − q
k
· exp (−p
k
t)) + q
k
· [1 − exp (−p
k
t)]
−
−
X
′
· P
k
· R
k
· exp [−p
k
t − q
k
exp (−p
k
t)]
X
2
·
h
1 + q
k
· [1 − exp (−p
k
t)] −
− p
k
t · [1 − q
k
exp (−p
k
t)] · (1 + q
k
· [1 − exp (−p
k
t)]) + p
k
· q
k
t · exp (−p
k
t)
i
.
Similarly, in the case of q
k
, we have
∂Π
BTC
∂q
k
=
∂
∂q
k
X
′
X
=
∂X
′
∂q
k
·
1
X
−
X
′
X
2
·
∂X
∂q
k
.
Here,
∂X
′
∂q
k
= −P
k
· R
k
· p
2
k
·
∂
∂q
k
exp [−p
k
t − q
k
exp (−p
k
t)] ×
×
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
=
= −P
k
· R
k
· p
2
k
·
"
∂
∂q
k
exp [−p
k
t − q
k
exp (−p
k
t)]
×
×
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
+
+ exp [−p
k
t − q
k
exp (−p
k
t)] ×
×
∂
∂q
k
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
#
=
= −P
k
· R
k
· p
2
k
·
"
exp [−p
k
t − q
k
exp (−p
k
t)] · (− exp (−p
k
t)) ×
×
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
+
+ exp [−p
k
t − q
k
exp (−p
k
t)] ×
×
2 · exp (−p
k
t) − [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) −
− q
k
· [1 − exp (−p
k
t)] · (− exp (−p
k
t))
#
= P
k
· R
k
· p
2
k
· exp (−p
k
t) · exp [−p
k
t − q
k
exp (−p
k
t)] ×
×
"
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
+
+ [exp (p
k
t) − 1] · (1 − q
k
· exp (−p
k
t)) + q
k
· [1 − exp (−p
k
t)] − 2
#
.
On the other hand,
∂X
∂q
k
= P
k
· R
k
· p
k
·
"
∂ exp [−p
k
t − q
k
exp (−p
k
t)]
∂q
k
· (1 + q
k
· [1 − exp (−p
k
t)]) +
+ exp [−p
k
t − q
k
exp (−p
k
t)] ·
∂
∂q
k
(1 + q
k
· [1 − exp (−p
k
t)])
#
=
= P
k
· R
k
· p
k
· exp [−p
k
t − q
k
exp (−p
k
t)] ×
× [− exp (−p
k
t) · (2 + q
k
· [1 − exp (−p
k
t)]) + 1] .
Therefore,
∂Π
BTC
∂q
k
=
P
k
· R
k
· p
2
k
· exp (−p
k
t) · exp [−p
k
t − q
k
exp (−p
k
t)]
X
×
×
"
2q
k
· exp (−p
k
t) − q
k
· [1 − exp (−p
k
t)] · (1 − q
k
· exp (−p
k
t)) − 1
+
+ [exp (p
k
t) − 1] · (1 − q
k
· exp (−p
k
t)) + q
k
· [1 − exp (−p
k
t)] − 2
#
−
−
X
′
· P
k
· R
k
· p
k
· exp [−p
k
t − q
k
exp (−p
k
t)]
X
2
×
× [− exp (−p
k
t) · (2 + q
k
· [1 − exp (−p
k
t)]) + 1]
Let us now proceed to the sensitivity derivatives calculations for ln P
BTC
. Apparently,
∂ ln P
BTC
∂p
k
=
∂ ln X
∂p
k
=
1
X
·
∂X
∂p
k
=
=
P
k
· R
k
· exp [−p
k
t − q
k
exp (−p
k
t)]
X
·
h
1 + q
k
· [1 − exp (−p
k
t)] −
− p
k
t · [1 − q
k
exp (−p
k
t)] · (1 + q
k
· [1 − exp (−p
k
t)]) + p
k
· q
k
t · exp (−p
k
t)
i
.
Similarly,
∂ ln P
BTC
∂q
k
=
∂ ln X
∂q
k
=
1
X
·
∂X
∂q
k
=
=
P
k
· R
k
· p
k
· exp [−p
k
t − q
k
exp (−p
k
t)]
X
×
× [− exp (−p
k
t) · (2 + q
k
· [1 − exp (−p
k
t)]) + 1] .
